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Comprehensive Guide on Schur's Triangulation Theorem

schedule Aug 12, 2023
Last updated
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Linear Algebra
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Recall from the section on orthogonal diagonalizationlink that if $\boldsymbol{A}$ is a symmetric matrix, then $\boldsymbol{A}$ can be orthogonally diagonalized as follows:

$$\boldsymbol{A}= \boldsymbol{QDQ}^T$$

Where:

In this section, we will look at a more relaxed case in which $\boldsymbol{A}$ is not necessarily symmetric but has $n$ real eigenvalues.

Theorem.

Triangulation theorem (Schur's theorem)

If $\boldsymbol{A}$ is an $n\times{n}$ matrix with $n$ real eigenvalueslink, then $\boldsymbol{A}$ can be expressed as:

$$\boldsymbol{A}= \boldsymbol{QUQ}^{T}$$

Where:

Note that this can be rewritten as $\boldsymbol{U}=\boldsymbol{Q}^T\boldsymbol{AQ}$ by propertylink of orthogonal matrices.

Proof. This proof is very similar to that of the real spectral theoremlink. We will prove this by induction on the size $n$ of the matrix $\boldsymbol{A}$. Consider the base case when $\boldsymbol{A}$ is an $1\times{1}$ matrix $\begin{pmatrix}a\end{pmatrix}$ with $1$ real eigenvalue. We now show that this eigenvalue must be equal to $a$ using the definitionlink of eigenvalues:

$$\begin{align*} \begin{pmatrix}a\end{pmatrix} \boldsymbol{x}&= \lambda\boldsymbol{x}\\ \begin{pmatrix}ax\end{pmatrix}&= \begin{pmatrix}\lambda{x}\end{pmatrix} \end{align*}$$

Where $\boldsymbol{x}$ is an eigenvector of $\boldsymbol{A}$ and $\lambda$ is the corresponding eigenvalue of $\boldsymbol{A}$. We can see that the equality holds only when we let $\lambda=a$. Now, matrix $\boldsymbol{A}$ can be written as:

$$\begin{pmatrix}a\end{pmatrix}= \begin{pmatrix}1\end{pmatrix} \begin{pmatrix}a\end{pmatrix} \begin{pmatrix}1\end{pmatrix}^T$$

Here, note the following:

  • $\begin{pmatrix}1\end{pmatrix}$ is trivially an orthogonal matrix by definitionlink because $\begin{pmatrix}1\end{pmatrix} \begin{pmatrix}1\end{pmatrix}^T= \begin{pmatrix}1\end{pmatrix} =\boldsymbol{I}$.

  • $\begin{pmatrix}a\end{pmatrix}$ on the right-hand side is trivially an upper-triangular matrix by definitionlink whose diagonal entry is the eigenvalue $a$.

Therefore, the theorem holds for the base case. We now assume that theorem holds for when $\boldsymbol{A}$ is an $(n-1)\times(n-1)$ matrix with $n-1$ real eigenvalues. Our goal is to show that the theorem holds for when $\boldsymbol{A}$ is an $n\times{n}$ matrix with $n$ real eigenvalues.

Suppose $\boldsymbol{A}$ is an $n\times{n}$ matrix with $n$ real eigenvalues $\lambda_1$, $ \lambda_2$, $\cdots$, $\lambda_n$. Let $\boldsymbol{x}_1$ be an unit eigenvector of $\boldsymbol{A}$. Let $\boldsymbol{q}_1= \boldsymbol{x}_1$ and use theoremlink to construct a basis for $\mathbb{R}^n$ like so:

$$\{\boldsymbol{q}_1,\boldsymbol{x}_2, \boldsymbol{x}_3,\cdots,\boldsymbol{x}_n\}$$

We now apply the Gram-Schmidt processlink to convert this basis into an orthonormal basislink:

$$\{\boldsymbol{q}_1,\boldsymbol{q}_2, \boldsymbol{q}_3,\cdots,\boldsymbol{q}_n\}$$

Remember, every vector in an orthonormal basis has unit length and is perpendicular to every other vector in the basis. Define an orthogonal matrixlink $\boldsymbol{Q}_1$ whose column vectors are these orthogonal basis vectors:

$$\begin{equation}\label{eq:BtgG8X9JZiQ50LCuF1Q} \boldsymbol{Q}_1= \begin{pmatrix} \vert&\vert&\cdots&\vert\\ \boldsymbol{q}_1&\boldsymbol{q}_2&\cdots&\boldsymbol{q}_n\\ \vert&\vert&\cdots&\vert \end{pmatrix} \end{equation}$$

Let matrix $\boldsymbol{B}=\boldsymbol{Q}^T_1\boldsymbol{AQ}_1$, which can be expressed as:

$$\begin{equation}\label{eq:hk06Dr36frpKSOynFnO} \begin{aligned}[b] \boldsymbol{B}&=\boldsymbol{Q}_1^T\boldsymbol{A}\boldsymbol{Q}_1 \\&= \begin{pmatrix}-&\boldsymbol{q}^T_1&-\\-&\boldsymbol{q}^T_2&-\\ \vdots&\vdots&\vdots\\-&\boldsymbol{q}^T_n&-\\ \end{pmatrix}\boldsymbol{A} \begin{pmatrix}\vert&\vert&\cdots&\vert\\ \boldsymbol{q}_1&\boldsymbol{q}_2&\cdots&\boldsymbol{q}_n\\ \vert&\vert&\cdots&\vert\end{pmatrix}\\ &= \begin{pmatrix}-&\boldsymbol{q}^T_1&-\\-&\boldsymbol{q}^T_2&-\\ \vdots&\vdots&\vdots\\-&\boldsymbol{q}^T_n&-\\ \end{pmatrix} \begin{pmatrix}\vert&\vert&\cdots&\vert\\ \boldsymbol{A}\boldsymbol{q}_1& \boldsymbol{A}\boldsymbol{q}_2&\cdots& \boldsymbol{A}\boldsymbol{q}_n\\ \vert&\vert&\cdots&\vert\end{pmatrix}\\ \end{aligned} \end{equation}$$

Here, we used theoremlink for the last step. The top-left entry of $\boldsymbol{B}$ is:

$$\begin{align*} \boldsymbol{q}^T_1\boldsymbol{Aq}_1 &=\boldsymbol{q}^T_1(\boldsymbol{Aq}_1)\\ &=\boldsymbol{q}^T_1(\lambda_1\boldsymbol{q}_1)\\ &=\lambda_1(\boldsymbol{q}^T_1\boldsymbol{q}_1)\\ &=\lambda_1(\boldsymbol{q}_1\cdot\boldsymbol{q}_1)\\ &=\lambda_1\Vert\boldsymbol{q}_1\Vert^2\\ &=\lambda_1(1)^2\\ &=\lambda_1 \end{align*}$$

Here, note the following:

  • the 2nd equality holds since $\boldsymbol{q}_1$ is an eigenvector of $\boldsymbol{A}$, we have that $\boldsymbol{Aq}_1=\lambda_1\boldsymbol{q}_1$ by definitionlink.

  • the 4th equality holds by propertylink of transpose.

  • the 5th equality holds by propertylink of dot product.

The other values in the first column are:

$$\begin{gather*} \boldsymbol{q}_2\cdot\boldsymbol{A}\boldsymbol{q}_1 \;{\color{blue}=}\;\boldsymbol{q}_2\cdot\lambda_1\boldsymbol{q}_1\;{\color{blue}=}\;\lambda_1(\boldsymbol{q}_2\cdot\boldsymbol{q}_1)\;{\color{blue}=}\;\lambda_1(0)\;{\color{blue}=}\;0\\ \boldsymbol{q}_3\cdot\boldsymbol{A}\boldsymbol{q}_1 \;{\color{blue}=}\;\boldsymbol{q}_3\cdot\lambda_1\boldsymbol{q}_1\;{\color{blue}=}\;\lambda_1(\boldsymbol{q}_3\cdot\boldsymbol{q}_1)\;{\color{blue}=}\;\lambda_1(0)\;{\color{blue}=}\;0\\ \vdots\\ \boldsymbol{q}_n\cdot\boldsymbol{A}\boldsymbol{q}_1 \;{\color{blue}=}\;\boldsymbol{q}_n\cdot\lambda_1\boldsymbol{q}_1 \;{\color{blue}=}\;\lambda_1(\boldsymbol{q}_n\cdot\boldsymbol{q}_1)\;{\color{blue}=}\;\lambda(0)\;{\color{blue}=}\;0 \\ \end{gather*}$$

This means that the first column has $\lambda_1$ as the first value and zeros for the rest. This allows us to view $\boldsymbol{B}$ as a block matrixlink like so:

$$\boldsymbol{B}= \begin{pmatrix} \lambda_1&\color{red}*&\color{red}*&\color{red}\cdots&\color{red}*\\ 0&\color{blue}*&\color{blue}*&\color{blue}\cdots&\color{blue}*\\ 0&\color{blue}*&\color{blue}*&\color{blue}\cdots&\color{blue}*\\ \vdots&\color{blue}\vdots&\color{blue}\vdots&\color{blue}\smash\ddots&\color{blue}\vdots\\ 0&\color{blue}*&\color{blue}*&\color{blue}\cdots&\color{blue}*\\ \end{pmatrix}= \begin{pmatrix} \lambda_1&\color{red}\boldsymbol{C}_1\\ \boldsymbol{0}&\color{blue}\boldsymbol{C}_2 \end{pmatrix}$$

Where:

  • $\boldsymbol{0}$ is an $(n-1)\times1$ zero matrix.

  • $\color{red}\boldsymbol{C}_1$ is some $1\times(n-1)$ matrix.

  • $\color{blue}\boldsymbol{C}_2$ is some $(n-1)\times(n-1)$ matrix.

Now, let's show that $\boldsymbol{C}_2$ has $n-1$ real eigenvalues. Recall that $\boldsymbol{B}=\boldsymbol{Q}^T_1\boldsymbol{AQ}$. Because $\boldsymbol{Q}^T_1=\boldsymbol{Q}^{-1}_1$ by propertylink of orthogonal matrices, we have that matrices $\boldsymbol{B}$ and $\boldsymbol{A}$ are similarlink by definition. By theoremlink, since $\boldsymbol{B}$ and $\boldsymbol{A}$ are similar, they must have the same eigenvalues. Since $\boldsymbol{A}$ is assumed to have $n$ real eigenvalues, $\boldsymbol{B}$ also has $n$ real eigenvalues $\lambda_1$, $\lambda_2$, $\cdots$, $\lambda_n$. This means that the characteristic polynomiallink of $\boldsymbol{B}$ is:

$$\begin{equation}\label{eq:O8gtxRbSzPTeRYUTM7s} \det(\boldsymbol{B}-\lambda\boldsymbol{I}_n)= (\lambda_1-\lambda)(\lambda_2-\lambda) (\lambda_3-\lambda) \cdots(\lambda_n-\lambda) \end{equation}$$

The left-hand side can also be written as:

$$\begin{equation}\label{eq:SWTYI1NOxzHcGUhWlpm} \begin{aligned}[b] \det(\boldsymbol{B}-\lambda\boldsymbol{I}_n)&= \det\left[\begin{pmatrix} \lambda_1&\color{red}*&\color{red}*&\color{red}\cdots&\color{red}*\\ 0&\color{blue}*&\color{blue}*&\color{blue}\cdots&\color{blue}*\\ 0&\color{blue}*&\color{blue}*&\color{blue}\cdots&\color{blue}*\\ \vdots&\color{blue}\vdots&\color{blue}\vdots&\color{blue}\smash\ddots&\color{blue}\vdots\\ 0&\color{blue}*&\color{blue}*&\color{blue}\cdots&\color{blue}*\\ \end{pmatrix}- \begin{pmatrix} \lambda&0&0&\cdots&0\\0&\lambda&0&\cdots&0\\0&0&\lambda&\cdots&0\\ \vdots&\vdots&\vdots&\smash\ddots&\vdots\\0&0&0&\cdots&\lambda\end{pmatrix}\right] \\&=\begin{vmatrix} \lambda_1-\lambda&\color{red}*&\color{red}*&\color{red}\cdots&\color{red}*\\ 0&{\color{blue}*}-\lambda&\color{blue}*&\color{blue}\cdots&\color{blue}*\\ 0&\color{blue}*&{\color{blue}*}-\lambda&\color{blue}\cdots&\color{blue}*\\ \vdots&\color{blue}\vdots&\color{blue}\vdots&\color{blue}\smash\ddots&\color{blue}\vdots\\ 0&\color{blue}*&\color{blue}*&\color{blue}\cdots&{\color{blue}*}-\lambda \end{vmatrix}\\&=(\lambda_1-\lambda)\begin{vmatrix}{\color{blue}*}-\lambda&\color{blue}*&\color{blue}\cdots&\color{blue}*\\ \color{blue}*&{\color{blue}*}-\lambda&\color{blue}\cdots&\color{blue}*\\\color{blue}\vdots&\color{blue}\vdots&\color{blue}\smash\ddots&\color{blue}\vdots\\ \color{blue}*&\color{blue}*&\color{blue}\cdots&{\color{blue}*}-\lambda\end{vmatrix}\\&=(\lambda_1-\lambda)\cdot\det\left[\;\;\begin{pmatrix} \color{blue}*&\color{blue}*&\color{blue}\cdots&\color{blue}*\\\color{blue}*&\color{blue}*&\color{blue}\cdots&\color{blue}*\\ \color{blue}\vdots&\color{blue}\vdots&\color{blue}\smash\ddots&\color{blue}\vdots\\ \color{blue}*&\color{blue}*&\color{blue}\cdots&\color{blue}* \end{pmatrix}-\begin{pmatrix}\lambda&0&\cdots&0\\0&\lambda&\cdots&0\\\vdots&\vdots&\smash\ddots&\vdots\\0&0&\cdots&\lambda\end{pmatrix}\;\;\right]\\ &=(\lambda_1-\lambda)\cdot\det({\color{blue}\boldsymbol{C_2}}-\lambda\boldsymbol{I}_{n-1}) \end{aligned} \end{equation}$$

Here, for the 3rd equality, we've computed the determinant by cofactor expansionlink along the first column.

Equating \eqref{eq:O8gtxRbSzPTeRYUTM7s} and \eqref{eq:SWTYI1NOxzHcGUhWlpm} gives us the characteristic polynomial of $\color{blue}\boldsymbol{C}_2$ below:

$$\begin{equation}\label{eq:AMsCbvLy3TIRiw0H2zT} \det({\color{blue}\boldsymbol{C}_2}- \lambda\boldsymbol{I}_{n-1})= (\lambda_2-\lambda) (\lambda_3-\lambda) \cdots(\lambda_n-\lambda) \end{equation}$$

Again, the roots of the characteristic polynomial of $\color{blue}\boldsymbol{C}_2$ are the eigenvalues of $\color{blue}\boldsymbol{C}_2$. This means that $\color{blue}\boldsymbol{C}_2$ has $n-1$ real eigenvalues $\lambda_2$, $\lambda_3$, $\cdots$, $\lambda_n$.

We now use our inductive assumption that any $(n-1)\times(n-1)$ matrix with $n-1$ real eigenvalues can be triangulated. This means that $\color{blue}\boldsymbol{C}_2$ can be triangulated like so:

$$\begin{equation}\label{eq:p6U3bidFVsaaYWgiy5M} \boldsymbol{R}=\boldsymbol{P}^T{\color{blue}\boldsymbol{C}_2}\boldsymbol{P} \end{equation}$$

Where:

Now, define an $n\times{n}$ matrix $\boldsymbol{Q}_2$ like so:

$$\boldsymbol{Q}_2= \begin{pmatrix} 1&\boldsymbol{0}_{1\times(n-1)}\\ \boldsymbol{0}_{(n-1)\times{1}}&\boldsymbol{P} \end{pmatrix}$$

Where:

  • $\boldsymbol{0}$ is a zero matrix - the subscripts are there to just emphasize their shape.

  • $\boldsymbol{P}$ is an $(n-1)\times(n-1)$ orthogonal matrix.

Let $\boldsymbol{Q}=\boldsymbol{Q}_1 \boldsymbol{Q}_2$. Our goal is to show that:

  • $\boldsymbol{Q}$ is orthogonal.

  • $\boldsymbol{Q}^T \boldsymbol{AQ}$ is upper triangular.

Let's first show that $\boldsymbol{Q}$ is orthogonal. By theoremlink, the transpose of $\boldsymbol{Q}_2$ is:

$$\boldsymbol{Q}_2^T= \begin{pmatrix} 1&\boldsymbol{0}\\ \boldsymbol{0}&\boldsymbol{P}^T \end{pmatrix}$$

Using the multiplicative property of block matriceslink, the product $\boldsymbol{Q}_2\boldsymbol{Q}^T_2$ is:

$$\begin{align*} \boldsymbol{Q}_2\boldsymbol{Q}_2^T&= \begin{pmatrix} 1&\boldsymbol{0}\\\boldsymbol{0}&\boldsymbol{P} \end{pmatrix} \begin{pmatrix} 1&\boldsymbol{0}\\\boldsymbol{0}&\boldsymbol{P}^T \end{pmatrix}\\ &= \begin{pmatrix} 1&\boldsymbol{0}\\\boldsymbol{0}&\boldsymbol{P}\boldsymbol{P}^T \end{pmatrix}\\ &= \begin{pmatrix} 1&\boldsymbol{0}\\\boldsymbol{0}&\boldsymbol{I}_{n-1} \end{pmatrix}\\ &=\boldsymbol{I}_{n} \end{align*}$$

Note that $\boldsymbol{PP}^T= \boldsymbol{I}_{n-1}$ because $\boldsymbol{P}$ is an $(n-1)\times(n-1)$ orthogonal matrix. Since $\boldsymbol{Q}_2\boldsymbol{Q}^T_2 = \boldsymbol{I}_n$, we have that $\boldsymbol{Q}_2$ is an orthogonal matrix by definitionlink. Recall that $\boldsymbol{Q}_1$ is defined as an orthogonal matrix in \eqref{eq:BtgG8X9JZiQ50LCuF1Q}, which means that $\boldsymbol{Q}=\boldsymbol{Q}_1\boldsymbol{Q}_2$ is also orthogonal since the product of two orthogonal matrices are orthogonal by theoremlink.

Next, let's show that $\boldsymbol{Q}^T\boldsymbol{AQ}$ is upper triangular:

$$\begin{equation}\label{eq:FrPrF6ZzjSesaaH3dp6} \begin{aligned}[b] \boldsymbol{Q}^T\boldsymbol{AQ}&= (\boldsymbol{Q}_1\boldsymbol{Q}_2)^T\boldsymbol{A}(\boldsymbol{Q}_1 \boldsymbol{Q}_2)\\&=\boldsymbol{Q}_2^T\boldsymbol{Q}_1^T\boldsymbol{A}\boldsymbol{Q}_1\boldsymbol{Q}_2\\&=\boldsymbol{Q}_2^T(\boldsymbol{Q}_1^T \boldsymbol{A}\boldsymbol{Q}_1)\boldsymbol{Q}_2\\&=\boldsymbol{Q}_2^T\boldsymbol{B}\boldsymbol{Q}_2\\ &=\begin{pmatrix}1&\boldsymbol{0}\\\boldsymbol{0}&\boldsymbol{P}^T\end{pmatrix}\begin{pmatrix}\lambda_1&\color{red}\boldsymbol{C}_1\\ \boldsymbol{0}&\color{blue}\boldsymbol{C}_2\end{pmatrix} \begin{pmatrix}1&\boldsymbol{0}\\\boldsymbol{0}&\boldsymbol{P}\end{pmatrix}\\ &=\begin{pmatrix}(1)(\lambda_1)+\boldsymbol{0}\boldsymbol{0}&(1){\color{red}\boldsymbol{C}_1}+\boldsymbol{0}{\color{blue}\boldsymbol{C}_2}\\ \boldsymbol{0}(\lambda_1)+\boldsymbol{P}^T\boldsymbol{0}&\boldsymbol{0}{\color{red}\boldsymbol{C}_1}+\boldsymbol{P}^T {\color{blue}\boldsymbol{C}_2}\end{pmatrix}\begin{pmatrix}1&\boldsymbol{0}\\\boldsymbol{0}&\boldsymbol{P} \end{pmatrix}\\&=\begin{pmatrix}\lambda_1&{\color{red}\boldsymbol{C}_1}\\\boldsymbol{0}&\boldsymbol{P}^T{\color{blue}\boldsymbol{C}_2} \end{pmatrix}\begin{pmatrix}1&\boldsymbol{0}\\\boldsymbol{0}&\boldsymbol{P} \end{pmatrix}\\&=\begin{pmatrix}(\lambda_1)(1)+{\color{red}\boldsymbol{C}_1}\boldsymbol{0} &(\lambda_1)\boldsymbol{0}+{\color{red}\boldsymbol{C}_1}\boldsymbol{P}\\ \boldsymbol{0}(1)+\boldsymbol{P}^T{\color{blue}\boldsymbol{C}_2}\boldsymbol{0} &\boldsymbol{0}\boldsymbol{0}+\boldsymbol{P}^T{\color{blue}\boldsymbol{C}_2}\boldsymbol{P}\\ \end{pmatrix}\\&=\begin{pmatrix}\lambda_1&{\color{red}\boldsymbol{C}_1}\boldsymbol{P}\\ \boldsymbol{0}&\boldsymbol{P}^T{\color{blue}\boldsymbol{C}_2}\boldsymbol{P} \end{pmatrix}\\&=\begin{pmatrix}\lambda_1&{\color{red}\boldsymbol{C}_1}\boldsymbol{P}\\ \boldsymbol{0}&\boldsymbol{R} \end{pmatrix} \end{aligned} \end{equation}$$

Here, note the following:

  • for the 2nd equality, we used propertylink of matrix transpose.

  • for the 4th equality, we used the definition $\boldsymbol{B}=\boldsymbol{Q}^T_1\boldsymbol{AQ}_1$.

  • to perform the multiplication of block matrices, we used theoremlink.

  • for the final equality, we used definition \eqref{eq:p6U3bidFVsaaYWgiy5M} where $\boldsymbol{R}$ is an upper triangular matrix whose diagonal entries are $\lambda_2$, $\lambda_3$, $\cdots$, $\lambda_n$.

Since \eqref{eq:FrPrF6ZzjSesaaH3dp6} is upper triangular, we have that $\boldsymbol{Q}^T\boldsymbol{AQ}$ is upper triangular. Moreover, the diagonal entries of $\boldsymbol{Q}^T\boldsymbol{AQ}$ are $\lambda_1$, $\lambda_2$, $\cdots$, $\lambda_n$, which are the eigenvalues of $\boldsymbol{A}$. By the principle of mathematical induction, the theorem holds for the general case. This completes the proof.

Although the triangulation theoremlink on its own does not have much practical importance, the theorem is useful for proving certain properties of eigenvalues and eigenvectors. We will demonstrate this later in the guide.

The triangulation theoremlink factorizes a matrix into orthogonal matrices and an upper triangular matrix. The next theorem shows that the upper triangular matrix can be replaced by a lower triangular matrix instead.

Theorem.

Lower triangulation theorem

If $\boldsymbol{A}$ is an $n\times{n}$ matrix with $n$ real eigenvalues, then $\boldsymbol{A}$ can be written as:

$$\boldsymbol{A}= \boldsymbol{QLQ}^{T}$$

Where:

Proof. By the triangulation theoremlink, if $\boldsymbol{A}$ is an $n\times{n}$ matrix with $n$ real eigenvalues, then $\boldsymbol{A}$ can be written as:

$$\boldsymbol{A}= \boldsymbol{QUQ}^{T}$$

Where:

  • $\boldsymbol{Q}$ is an $n\times{n}$ orthogonal matrix.

  • $\boldsymbol{U}$ is an $n\times{n}$ upper triangular matrix.

Taking the transpose of both sides gives:

$$\begin{align*} \boldsymbol{A}^T&= (\boldsymbol{QUQ}^T)^T\\ &=(\boldsymbol{Q}^T)^T\boldsymbol{U}^T\boldsymbol{Q}^T\\ &=\boldsymbol{Q}\boldsymbol{L}\boldsymbol{Q}^T\\ \end{align*}$$

Here, note the following:

  • the second equality holds by propertylink of matrix transpose.

  • $\boldsymbol{U}^T$ is equal to some lower triangular matrix $\boldsymbol{L}$ by theoremlink.

This completes the proof.

We will now use the triangulation theoremlink to prove two important properties of eigenvalues.

Theorem.

Determinant of a matrix whose eigenvalues are all real is equal to the product of all eigenvalues

If $\boldsymbol{A}$ is an $n\times{n}$ matrix with real eigenvalues $\lambda_1$, $\lambda_2$, $\cdots$, $\lambda_n$, then the determinantlink of $\boldsymbol{A}$ is:

$$\det(\boldsymbol{A})= \prod^n_{i=1}\lambda_i$$

Note that the eigenvalues may not necessarily be all unique.

Proof. By the triangulation theoremlink, if $\boldsymbol{A}$ is an $n\times{n}$ matrix with $n$ real eigenvalues $\lambda_1$, $\lambda_2$, $\cdots$, $\lambda_n$, then $\boldsymbol{A}$ can be expressed as:

$$\begin{equation}\label{eq:KVLJvXjGOB3d9Kpu732} \boldsymbol{A}= \boldsymbol{QUQ}^{T} \end{equation}$$

Where:

  • $\boldsymbol{Q}$ is an $n\times{n}$ orthogonal matrix.

  • $\boldsymbol{U}$ is an $n\times{n}$ upper triangular matrix whose diagonal entries are the eigenvalues of $\boldsymbol{A}$.

Taking the determinant of both sides of \eqref{eq:KVLJvXjGOB3d9Kpu732} gives:

$$\begin{align*} \det(\boldsymbol{A})&= \det(\boldsymbol{QUQ}^{T})\\ &= \det(\boldsymbol{Q})\cdot \det(\boldsymbol{U})\cdot \det(\boldsymbol{Q}^T)\\ &=\det(\boldsymbol{Q})\cdot\det(\boldsymbol{Q}^T)\cdot \det(\boldsymbol{U})\\ &=\det(\boldsymbol{QQ}^T)\cdot\det(\boldsymbol{U})\\ &=\det(\boldsymbol{I})\cdot\det(\boldsymbol{U})\\ &=(1)\cdot\det(\boldsymbol{U})\\ &=\det(\boldsymbol{U}) \end{align*}$$

Here, we've used the following properties:

Now, by propertylink of triangular matrices, the determinant of $\boldsymbol{U}$ is equal to the product of its diagonal entries. The diagonal entries of $\boldsymbol{U}$ are the eigenvalues of $\boldsymbol{A}$ so:

$$\begin{align*} \det(\boldsymbol{A})&= \det(\boldsymbol{U})\\ &=\lambda_1\lambda_2\cdots\lambda_n\\ &=\prod^n_{i=1}\lambda_i \end{align*}$$

This completes the proof.

Theorem.

Trace of a matrix whose eigenvalues are all real is equal to the sum of all eigenvalues

If $\boldsymbol{A}$ is an $n\times{n}$ matrix with real eigenvalues $\lambda_1$, $\lambda_2$, $\cdots$, $\lambda_n$, then the tracelink of $\boldsymbol{A}$ is:

$$\mathrm{tr}(\boldsymbol{A})= \sum^n_{i=1}\lambda_i$$

Note that the eigenvalues may not necessarily be all unique.

Proof. By the triangulation theoremlink, if $\boldsymbol{A}$ is an $n\times{n}$ matrix with $n$ real eigenvalues $\lambda_1$, $\lambda_2$, $\cdots$, $\lambda_n$, then $\boldsymbol{A}$ can be expressed as:

$$\begin{equation}\label{eq:hXsvzcGON9llfAMLWn7} \boldsymbol{A}= \boldsymbol{QUQ}^{T} \end{equation}$$

Where:

  • $\boldsymbol{Q}$ is an $n\times{n}$ orthogonal matrix.

  • $\boldsymbol{U}$ is an $n\times{n}$ upper triangular matrix whose diagonal entries are the eigenvalues of $\boldsymbol{A}$.

Taking the trace of both sides of \eqref{eq:hXsvzcGON9llfAMLWn7} gives:

$$\begin{align*} \mathrm{tr}(\boldsymbol{A})&= \mathrm{tr}(\boldsymbol{QUQ}^{T})\\ &=\mathrm{tr}\big((\boldsymbol{QU})\boldsymbol{Q}^{T}\big)\\ &=\mathrm{tr}\big(\boldsymbol{Q}^{T}(\boldsymbol{QU})\big)\\ &=\mathrm{tr}\big(\boldsymbol{Q}^{T}\boldsymbol{QU}\big)\\ &=\mathrm{tr}\big(\boldsymbol{I}\boldsymbol{U}\big)\\ &=\mathrm{tr}\big(\boldsymbol{U}\big)\\ &=\lambda_1+\lambda_2+\cdots+\lambda_n\\ &=\sum^n_{i=1}\lambda_i \end{align*}$$

Here, we used the propertylink of trace that $\mathrm{tr}(\boldsymbol{CD})= \mathrm{tr}(\boldsymbol{DC})$ for any matrices $\boldsymbol{C}$ and $\boldsymbol{D}$. This completes the proof.

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Published by Isshin Inada
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