Definition.
Diagonal matrix
A diagonal matrix is a square matrix whose non-diagonal entries are all zero. For instance, here is a $3\times3$ diagonal matrix:
$$\boldsymbol{D}=\begin{pmatrix}
2&0&0\\
0&6&0\\
0&0&4
\end{pmatrix}$$
Diagonal matrices are usually denoted by a bold uppercase letter $\boldsymbol{D}$.
Example.
Identity matrix
The identity matrix $\boldsymbol{I}_n$ is a classic example of a diagonal matrix. Here's the $3\times3$ identity matrix:
$$\boldsymbol{I}_3=\begin{pmatrix}
1&0&0\\
0&1&0\\
0&0&1
\end{pmatrix}$$
Theorem.
Transpose of a diagonal matrix equals itself
If $\boldsymbol{D}$ is a diagonal matrix, then:
$$\boldsymbol{D}^T=\boldsymbol{D}$$
Proof. By theoremlink, taking the transpose of a square matrix does not change the diagonal entries. Since diagonal matrices are square matrices whose non-diagonal entries are all zero by definitionlink, we conclude that the transpose of a diagonal matrix is itself. This completes the proof.
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Theorem.
Product of a matrix and a diagonal matrix
Consider an $m\times{n}$ matrix $\boldsymbol{A}$ and an $n\times{n}$ diagonal matrix $\boldsymbol{D}$ below:
$$\boldsymbol{A}
=
\begin{pmatrix}
\vert&\vert&\cdots&\vert\\
\boldsymbol{a_1}&\boldsymbol{a_2}&\cdots&\boldsymbol{a_n}\\
\vert&\vert&\cdots&\vert\\
\end{pmatrix},\;\;\;\;\;
\boldsymbol{D}=
\begin{pmatrix}
d_{11}&0&\cdots&0\\
0&d_{22}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&d_{nn}\\
\end{pmatrix}$$
Where the columns of matrix $\boldsymbol{A}$ are represented by vectors $\boldsymbol{a}_1$, $\boldsymbol{a}_2$, $\cdots$, $\boldsymbol{a}_n$.
The product $\boldsymbol{AD}$ is:
$$\boldsymbol{AD}
=
\begin{pmatrix}
\vert&\vert&\cdots&\vert\\
d_{11}\boldsymbol{a_1}&
d_{22}\boldsymbol{a_2}
&\cdots&
d_{nn}\boldsymbol{a_n}\\
\vert&\vert&\cdots&\vert\\
\end{pmatrix}$$
Proof. Let matrix $\boldsymbol{A}$ be represented as:
$$\boldsymbol{A}=
\begin{pmatrix}
\vert&\vert&\cdots&\vert\\
\boldsymbol{a_1}&\boldsymbol{a_2}&\cdots&\boldsymbol{a_n}\\
\vert&\vert&\cdots&\vert\\
\end{pmatrix}=
\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}
\end{pmatrix}$$
The product $\boldsymbol{AD}$ is:
$$\begin{align*}
\boldsymbol{AD}&=\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\smash\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}
\end{pmatrix}
\begin{pmatrix}
d_{11}&0&\cdots&0\\
0&d_{22}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&d_{nn}\\
\end{pmatrix}\\
&=\begin{pmatrix}
a_{11}d_{11}&a_{12}d_{22}&\cdots&a_{1n}d_{nn}\\
a_{21}d_{11}&a_{22}d_{22}&\cdots&a_{2n}d_{nn}\\
\vdots&\vdots&\smash\ddots&\vdots\\
a_{m1}d_{11}&a_{m2}d_{22}&\cdots&a_{mn}d_{nn}
\end{pmatrix}\\
&=\begin{pmatrix}
\vert&\vert&\cdots&\vert\\
d_{11}\boldsymbol{a}_1&
d_{22}\boldsymbol{a}_2
&\cdots&
d_{nn}\boldsymbol{a}_n\\
\vert&\vert&\cdots&\vert\\
\end{pmatrix}
\end{align*}$$
This completes the proof.
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Example.
Computing the product of a matrix and a diagonal matrix
Compute the following matrix product:
$$\begin{pmatrix}
1&4\\
5&6\\
\end{pmatrix}
\begin{pmatrix}
2&0\\
0&3\\
\end{pmatrix}$$
Solution. Let $\boldsymbol{A}$ denote the left matrix. We multiply the first column of $\boldsymbol{A}$ by $2$ and the second column by $3$ to get:
$$\begin{pmatrix}
1&4\\
5&6\\
\end{pmatrix}
\begin{pmatrix}
2&0\\
0&3\\
\end{pmatrix}=
\begin{pmatrix}
2&12\\
10&18\\
\end{pmatrix}$$
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Theorem.
Product of a diagonal matrix and a matrix
Consider an $n\times{m}$ matrix $\boldsymbol{A}$ and an $n\times{n}$ diagonal matrix $\boldsymbol{D}$ below:
$$\boldsymbol{A}=\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1m}\\
a_{21}&a_{22}&\cdots&a_{2m}\\
\vdots&\vdots&\ddots&\vdots\\
a_{n1}&a_{n2}&\cdots&a_{nm}
\end{pmatrix}=
\begin{pmatrix}
-&\boldsymbol{a}_1&-\\
-&\boldsymbol{a}_2&-\\
\vdots&\vdots&\vdots\\
-&\boldsymbol{a}_n&-\\
\end{pmatrix}
,\;\;\;\;\;\;
\boldsymbol{D}=\begin{pmatrix}
d_{11}&0&\cdots&0\\
0&d_{22}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&d_{nn}\\
\end{pmatrix}$$
Where $\boldsymbol{A}$ is represented as a collection of row vectors.
The product $\boldsymbol{DA}$ is:
$$\boldsymbol{DA}=
\begin{pmatrix}
-&d_{11}\boldsymbol{a}_1&-\\
-&d_{22}\boldsymbol{a}_2&-\\
\vdots&\vdots&\vdots\\
-&d_{nn}\boldsymbol{a}_n&-\\
\end{pmatrix}$$
Proof. Let $\boldsymbol{A}$ be an $m\times{n}$ matrix represented as row vectors and $\boldsymbol{D}$ be a diagonal matrix:
$$\begin{align*}
\boldsymbol{DA}&=
\begin{pmatrix}
d_{11}&0&\cdots&0\\
0&d_{22}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&d_{nn}\\
\end{pmatrix}\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1m}\\
a_{21}&a_{22}&\cdots&a_{2m}\\
\vdots&\vdots&\smash\ddots&\vdots\\
a_{n1}&a_{n2}&\cdots&a_{nm}
\end{pmatrix}\\
&=\begin{pmatrix}
d_{11}a_{11}&d_{11}a_{12}&\cdots&d_{11}a_{1m}\\
d_{22}a_{21}&d_{22}a_{22}&\cdots&d_{22}a_{2m}\\
\vdots&\vdots&\ddots&\vdots\\
d_{nn}a_{n1}&d_{nn}a_{n2}&\cdots&d_{nn}a_{nm}
\end{pmatrix}\\
&=\begin{pmatrix}
-&d_{11}\boldsymbol{a}_1&-\\
-&d_{22}\boldsymbol{a}_2&-\\
\vdots&\vdots&\vdots\\
-&d_{nn}\boldsymbol{a}_n&-\\
\end{pmatrix}
\end{align*}$$
This completes the proof.
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Example.
Finding the product of a diagonal matrix and a matrix
Compute the following matrix product:
$$\begin{pmatrix}
3&0&0\\0&2&0\\0&0&1
\end{pmatrix}
\begin{pmatrix}
5&2&4\\6&3&1\\1&0&2
\end{pmatrix}$$
Proof. The matrix product is:
$$\begin{align*}
\begin{pmatrix}
3&0&0\\0&2&0\\0&0&1
\end{pmatrix}
\begin{pmatrix}
5&2&4\\6&3&1\\1&0&2
\end{pmatrix}&=
\begin{pmatrix}
(3)5&(3)2&(3)4\\(2)6&(2)3&(2)1\\
(1)1&(1)0&(1)2
\end{pmatrix}\\
&=
\begin{pmatrix}
15&6&12\\12&6&2\\
1&0&2
\end{pmatrix}
\end{align*}$$
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Theorem.
Taking the power of diagonal matrices
Consider the following $n\times{n}$ diagonal matrix:
$$\boldsymbol{D}=\begin{pmatrix}
d_{11}&0&\cdots&0\\
0&d_{22}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&d_{nn}\\
\end{pmatrix}$$
Raising $\boldsymbol{D}$ to the power of some positive integer $k$ involves raising the diagonal entries to the power of $k$, that is:
$$\boldsymbol{D}^k=\begin{pmatrix}
d_{11}^k&0&\cdots&0\\
0&d_{22}^k&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&d_{nn}^k\\
\end{pmatrix}$$
Proof. We prove this by induction. Consider the base case when $k=1$, which is trivially true:
$$\boldsymbol{D}^1=
\boldsymbol{D}=\begin{pmatrix}
d_{11}&0&\cdots&0\\
0&d_{22}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&d_{nn}\\
\end{pmatrix}=\begin{pmatrix}
d_{11}^1&0&\cdots&0\\
0&d_{22}^1&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&d_{nn}^1\\
\end{pmatrix}$$
We now assume the theorem holds when the power is raised to $k-1$, that is:
$$\begin{equation}\label{eq:xWOtz2qPTDfaXzfTIsH}
\boldsymbol{D}^{k-1}=\begin{pmatrix}
d_{11}^{k-1}&0&\cdots&0\\
0&d_{22}^{k-1}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&d_{nn}^{k-1}
\end{pmatrix}
\end{equation}$$
Our goal is to show that the theorem holds when the power is raised to $k$. This is quite easy because:
$$\begin{align*}
\boldsymbol{D}^k&=
\boldsymbol{D}^{k-1}\boldsymbol{D}
\end{align*}$$
We now use the inductive assumption \eqref{eq:xWOtz2qPTDfaXzfTIsH} to get:
$$\begin{align*}
\boldsymbol{D}^k
&=\boldsymbol{D}^{k-1}\boldsymbol{D}\\
&=\begin{pmatrix}
d_{11}^{k-1}&0&\cdots&0\\
0&d_{22}^{k-1}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&d_{nn}^{k-1}
\end{pmatrix}
\begin{pmatrix}
d_{11}&0&\cdots&0\\
0&d_{22}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&d_{nn}
\end{pmatrix}\\
&=
\begin{pmatrix}
d_{11}^k&0&\cdots&0\\
0&d_{22}^k&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&d_{nn}^k
\end{pmatrix}
\end{align*}$$
By the principle of mathematical induction, the theorem holds for the general case. This completes the proof.
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Example.
Computing the power of an 2x2 matrix
Consider the following diagonal matrix:
$$\boldsymbol{D}=\begin{pmatrix}
2&0\\
0&1\\
\end{pmatrix}$$
Compute $\boldsymbol{D}^3$.
Solution. $\boldsymbol{D}^3$ can easily be computed by raising each diagonal entry to the power of $3$ like so:
$$\begin{align*}
\boldsymbol{D}^3&=
\begin{pmatrix}
2^3&0\\
0&1^3\\
\end{pmatrix}\\
&=
\begin{pmatrix}
8&0\\
0&1\\
\end{pmatrix}
\end{align*}$$
This is a very neat property of diagonal matrices because taking powers of numbers is computationally much cheaper than matrix multiplication!
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Theorem.
Diagonal matrix is a triangular matrix
If $\boldsymbol{D}$ is a diagonal matrix, then $\boldsymbol{D}$ is both a lower and upper triangular matrix.
Proof. Diagonal matrix $\boldsymbol{D}$ is a lower triangular matrix because all the values above the diagonal entries are zero. Similarly, $\boldsymbol{D}$ is also an upper triangular matrix because all the values below the diagonal entries are zero. This completes the proof.
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Theorem.
Determinant of a diagonal matrix is equal to the product of its diagonal entries
If $\boldsymbol{D}$ is a diagonal matrix, then the determinant of $\boldsymbol{D}$ is equal to the product of its diagonal entries.
Proof. Since a diagonal matrix is triangular, theoremlink applies. This completes the proof.
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Theorem.
Diagonal matrix is invertible if and only if every diagonal entry is non-zero
Let $\boldsymbol{D}$ be a diagonal matrix. $\boldsymbol{D}$ is invertiblelink if and only if every diagonal entry of $\boldsymbol{D}$ is non-zero.
Proof. Since a diagonal matrix is triangular, theoremlink applies. This completes the proof.
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Theorem.
Product of two diagonal matrices is also diagonal
If $\boldsymbol{D}_1$ and $\boldsymbol{D}_2$ are $n\times{n}$ diagonal matrices, then their product $\boldsymbol{D}_1\boldsymbol{D}_2$ is a diagonal matrix whose diagonal entry contains pairwise products of the diagonal entries of $\boldsymbol{D}_1$ and $\boldsymbol{D}_2$.
Proof. Let $\boldsymbol{D}_1$ and $\boldsymbol{D}_2$ be the following diagonal matrices:
$$\boldsymbol{D}_1=
\begin{pmatrix}
a_{11}&0&\cdots&0\\
0&a_{22}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&a_{nn}
\end{pmatrix},\;\;\;\;\;\;
\boldsymbol{D}_2=
\begin{pmatrix}
b_{11}&0&\cdots&0\\
0&b_{22}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&b_{nn}
\end{pmatrix}$$
The product $\boldsymbol{D}_1\boldsymbol{D}_2$ is:
$$\begin{align*}
\boldsymbol{D}_1\boldsymbol{D}_2&=
\begin{pmatrix}
a_{11}&0&\cdots&0\\
0&a_{22}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&a_{nn}
\end{pmatrix}
\begin{pmatrix}
b_{11}&0&\cdots&0\\0&b_{22}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&b_{nn}
\end{pmatrix}\\
&=
\begin{pmatrix}
a_{11}b_{11}&0&\cdots&0\\0&a_{22}b_{22}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&a_{nn}b_{nn}
\end{pmatrix}
\end{align*}$$
This completes the proof.
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Theorem.
Product of a triangular matrix and a diagonal matrix
Let $\boldsymbol{A}$ be a triangular matrix and $\boldsymbol{D}$ be a diagonal matrix:
if $\boldsymbol{A}$ is upper triangular, then $\boldsymbol{AD}$ and $\boldsymbol{DA}$ are upper triangular matrices.
if $\boldsymbol{A}$ is lower triangular, then $\boldsymbol{AD}$ and $\boldsymbol{DA}$ are lower triangular matrix.
Note that the diagonals of $\boldsymbol{AD}$ and $\boldsymbol{DA}$ are equal to the pairwise products of the diagonal entries of $\boldsymbol{A}$ and $\boldsymbol{D}$.
Proof. Let $\boldsymbol{A}$ be an upper triangular matrix and $\boldsymbol{D}$ be a diagonal matrix. $\boldsymbol{D}$ is also an upper triangular matrix by theoremlink, which means that the product $\boldsymbol{AD}$ is upper triangular by theoremlink. We can also apply the same theoremlink to conclude that $\boldsymbol{DA}$ is upper triangular. Finally, by theoremlink, the diagonal entries of $\boldsymbol{AD}$ and $\boldsymbol{DA}$ will be the pairwise products of the diagonal entries of $\boldsymbol{A}$ and $\boldsymbol{D}$. The proof for the lower triangular case is analogous. This completes the proof.
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Theorem.
Inverse of a diagonal matrix
Let $\boldsymbol{D}$ be a diagonal matrix:
$$\boldsymbol{D}=
\begin{pmatrix}
d_{11}&0&\cdots&0\\
0&d_{22}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&d_{nn}
\end{pmatrix}$$
If every diagonal entry of $\boldsymbol{D}$ is non-zero, then $\boldsymbol{D}^{-1}$ is computed by:
$$\boldsymbol{D}^{-1}=
\begin{pmatrix}
\frac{1}{d_{11}}&0&\cdots&0\\
0&\frac{1}{d_{22}}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&\frac{1}{d_{nn}}
\end{pmatrix}$$
This also means that the inverse of a diagonal matrix is also diagonal.
Proof. Suppose we have an $n\times{n}$ diagonal matrix $\boldsymbol{D}$ and another matrix $\boldsymbol{A}$ below:
$$\begin{align*}
\boldsymbol{D}=
\begin{pmatrix}
d_{11}&0&\cdots&0\\
0&d_{22}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&d_{nn}
\end{pmatrix},\;\;\;\;\;\;\;
\boldsymbol{A}=\begin{pmatrix}
\frac{1}{d_{11}}&0&\cdots&0\\
0&\frac{1}{d_{22}}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&\frac{1}{d_{nn}}
\end{pmatrix}
\end{align*}$$
Our goal is to show that $\boldsymbol{A}=\boldsymbol{D}^{-1}$. The product $\boldsymbol{DA}$ is:
$$\begin{align*}
\boldsymbol{D}\boldsymbol{A}=
\begin{pmatrix}
d_{11}&0&\cdots&0\\
0&d_{22}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&d_{nn}
\end{pmatrix}\begin{pmatrix}
\frac{1}{d_{11}}&0&\cdots&0\\
0&\frac{1}{d_{22}}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&\frac{1}{d_{nn}}
\end{pmatrix}
\end{align*}$$
By theoremlink, taking the product of two diagonal matrices involves multiplying the corresponding diagonal entries:
$$\boldsymbol{D}\boldsymbol{A}=
\begin{pmatrix}
\frac{d_{11}}{d_{11}}&0&\cdots&0\\
0&\frac{d_{22}}{d_{22}}&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&\frac{d_{nn}}{d_{nn}}
\end{pmatrix}=
\begin{pmatrix}
1&0&\cdots&0\\
0&1&\cdots&0\\
\vdots&\vdots&\smash\ddots&\vdots\\
0&0&\cdots&1
\end{pmatrix}=\boldsymbol{I}_n
$$
Because $\boldsymbol{DA}=\boldsymbol{I}_n$, we have that $\boldsymbol{D}^{-1}=\boldsymbol{A}$ by definitionlink of inverse matrices. This completes the proof.
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Example.
Finding the inverse of a diagonal matrix
Find the inverse of the following diagonal matrix:
$$\boldsymbol{D}=
\begin{pmatrix}
3&0&0\\
0&2&0\\
0&0&1\\
\end{pmatrix}$$
Solution. The inverse of $\boldsymbol{D}$ is a diagonal matrix whose diagonal entries are the reciprocal:
$$\boldsymbol{D}^{-1}=
\begin{pmatrix}
1/3&0&0\\
0&1/2&0\\
0&0&1\\
\end{pmatrix}$$
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