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 $A$ and an $n \times n$ diagonal matrix $D$ below:

A = (\begin{matrix} | & | & \dots & | \\ a_{1} & a_{2} & \dots & a_{n} \\ | & | & \dots & | \end{matrix}), D = (\begin{matrix} d_{11} & 0 & \dots & 0 \\ 0 & d_{22} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & d_{n n} \end{matrix})

Where the columns of matrix $A$ are represented by vectors $a_{1}$ , $a_{2}$ , $\dots$ , $a_{n}$ .

The product $A D$ is:

A D = (\begin{matrix} | & | & \dots & | \\ d_{11} a_{1} & d_{22} a_{2} & \dots & d_{n n} a_{n} \\ | & | & \dots & | \end{matrix})

Proof. Let matrix $A$ be represented as:

A = (\begin{matrix} | & | & \dots & | \\ a_{1} & a_{2} & \dots & a_{n} \\ | & | & \dots & | \end{matrix}) = (\begin{matrix} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{m 1} & a_{m 2} & \dots & a_{m n} \end{matrix})

The product $A D$ is:

\begin{aligned} A D & = (\begin{array}{c} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{m 1} & a_{m 2} & \dots & a_{m n} \end{array}) (\begin{array}{c} d_{11} & 0 & \dots & 0 \\ 0 & d_{22} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & d_{n n} \end{array}) \\ = (\begin{array}{c} a_{11} d_{11} & a_{12} d_{22} & \dots & a_{1 n} d_{n n} \\ a_{21} d_{11} & a_{22} d_{22} & \dots & a_{2 n} d_{n n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{m 1} d_{11} & a_{m 2} d_{22} & \dots & a_{m n} d_{n n} \end{array}) \\ = (\begin{array}{c} | & | & \dots & | \\ d_{11} a_{1} & d_{22} a_{2} & \dots & d_{n n} a_{n} \\ | & | & \dots & | \end{array}) \end{aligned}

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{matrix} 1 & 4 \\ 5 & 6 \end{matrix}) (\begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix})

Solution. Let $A$ denote the left matrix. We multiply the first column of $A$ by $2$ and the second column by $3$ to get:

(\begin{matrix} 1 & 4 \\ 5 & 6 \end{matrix}) (\begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix}) = (\begin{matrix} 2 & 12 \\ 10 & 18 \end{matrix})

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Theorem.

Product of a diagonal matrix and a matrix

Consider an $n \times m$ matrix $A$ and an $n \times n$ diagonal matrix $D$ below:

A = (\begin{matrix} a_{11} & a_{12} & \dots & a_{1 m} \\ a_{21} & a_{22} & \dots & a_{2 m} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{n 1} & a_{n 2} & \dots & a_{n m} \end{matrix}) = (\begin{matrix} - & a_{1} & - \\ - & a_{2} & - \\ ⋮ & ⋮ & ⋮ \\ - & a_{n} & - \end{matrix}), D = (\begin{matrix} d_{11} & 0 & \dots & 0 \\ 0 & d_{22} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & d_{n n} \end{matrix})

Where $A$ is represented as a collection of row vectors.

The product $D A$ is:

D A = (\begin{matrix} - & d_{11} a_{1} & - \\ - & d_{22} a_{2} & - \\ ⋮ & ⋮ & ⋮ \\ - & d_{n n} a_{n} & - \end{matrix})

Proof. Let $A$ be an $m \times n$ matrix represented as row vectors and $D$ be a diagonal matrix:

\begin{aligned} D A & = (\begin{array}{c} d_{11} & 0 & \dots & 0 \\ 0 & d_{22} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & d_{n n} \end{array}) (\begin{array}{c} a_{11} & a_{12} & \dots & a_{1 m} \\ a_{21} & a_{22} & \dots & a_{2 m} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{n 1} & a_{n 2} & \dots & a_{n m} \end{array}) \\ = (\begin{array}{c} d_{11} a_{11} & d_{11} a_{12} & \dots & d_{11} a_{1 m} \\ d_{22} a_{21} & d_{22} a_{22} & \dots & d_{22} a_{2 m} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ d_{n n} a_{n 1} & d_{n n} a_{n 2} & \dots & d_{n n} a_{n m} \end{array}) \\ = (\begin{array}{c} - & d_{11} a_{1} & - \\ - & d_{22} a_{2} & - \\ ⋮ & ⋮ & ⋮ \\ - & d_{n n} a_{n} & - \end{array}) \end{aligned}

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{matrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} 5 & 2 & 4 \\ 6 & 3 & 1 \\ 1 & 0 & 2 \end{matrix})

Proof. The matrix product is:

\begin{aligned} (\begin{array}{c} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{array}) (\begin{array}{c} 5 & 2 & 4 \\ 6 & 3 & 1 \\ 1 & 0 & 2 \end{array}) & = (\begin{array}{c} (3) 5 & (3) 2 & (3) 4 \\ (2) 6 & (2) 3 & (2) 1 \\ (1) 1 & (1) 0 & (1) 2 \end{array}) \\ = (\begin{array}{c} 15 & 6 & 12 \\ 12 & 6 & 2 \\ 1 & 0 & 2 \end{array}) \end{aligned}

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Theorem.

Taking the power of diagonal matrices

Consider the following $n \times n$ diagonal matrix:

D = (\begin{matrix} d_{11} & 0 & \dots & 0 \\ 0 & d_{22} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & d_{n n} \end{matrix})

Raising $D$ to the power of some positive integer $k$ involves raising the diagonal entries to the power of $k$ , that is:

D^{k} = (\begin{matrix} d_{11}^{k} & 0 & \dots & 0 \\ 0 & d_{22}^{k} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & d_{n n}^{k} \end{matrix})

Proof. We prove this by induction. Consider the base case when $k = 1$ , which is trivially true:

D^{1} = D = (\begin{matrix} d_{11} & 0 & \dots & 0 \\ 0 & d_{22} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & d_{n n} \end{matrix}) = (\begin{matrix} d_{11}^{1} & 0 & \dots & 0 \\ 0 & d_{22}^{1} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & d_{n n}^{1} \end{matrix})

We now assume the theorem holds when the power is raised to $k - 1$ , that is:

\begin{matrix} (1) & D^{k - 1} = (\begin{matrix} d_{11}^{k - 1} & 0 & \dots & 0 \\ 0 & d_{22}^{k - 1} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & d_{n n}^{k - 1} \end{matrix}) \end{matrix}

Our goal is to show that the theorem holds when the power is raised to $k$ . This is quite easy because:

\begin{aligned} D^{k} & = D^{k - 1} D \end{aligned}

We now use the inductive assumption $(1)$ to get:

\begin{aligned} D^{k} & = D^{k - 1} D \\ = (\begin{array}{c} d_{11}^{k - 1} & 0 & \dots & 0 \\ 0 & d_{22}^{k - 1} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & d_{n n}^{k - 1} \end{array}) (\begin{array}{c} d_{11} & 0 & \dots & 0 \\ 0 & d_{22} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & d_{n n} \end{array}) \\ = (\begin{array}{c} d_{11}^{k} & 0 & \dots & 0 \\ 0 & d_{22}^{k} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & d_{n n}^{k} \end{array}) \end{aligned}

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:

D = (\begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix})

Compute $D^{3}$ .

Solution. $D^{3}$ can easily be computed by raising each diagonal entry to the power of $3$ like so:

\begin{aligned} D^{3} & = (\begin{array}{c} 2^{3} & 0 \\ 0 & 1^{3} \end{array}) \\ = (\begin{array}{c} 8 & 0 \\ 0 & 1 \end{array}) \end{aligned}

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 $D$ is a diagonal matrix, then $D$ is both a lower and upper triangular matrix.

Proof. Diagonal matrix $D$ is a lower triangular matrix because all the values above the diagonal entries are zero. Similarly, $D$ 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 $D$ is a diagonal matrix, then the determinant of $D$ 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 $D$ be a diagonal matrix. $D$ is invertiblelink if and only if every diagonal entry of $D$ 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 $D_{1}$ and $D_{2}$ are $n \times n$ diagonal matrices, then their product $D_{1} D_{2}$ is a diagonal matrix whose diagonal entry contains pairwise products of the diagonal entries of $D_{1}$ and $D_{2}$ .

Proof. Let $D_{1}$ and $D_{2}$ be the following diagonal matrices:

D_{1} = (\begin{matrix} a_{11} & 0 & \dots & 0 \\ 0 & a_{22} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & a_{n n} \end{matrix}), D_{2} = (\begin{matrix} b_{11} & 0 & \dots & 0 \\ 0 & b_{22} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & b_{n n} \end{matrix})

The product $D_{1} D_{2}$ is:

\begin{aligned} D_{1} D_{2} & = (\begin{array}{c} a_{11} & 0 & \dots & 0 \\ 0 & a_{22} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & a_{n n} \end{array}) (\begin{array}{c} b_{11} & 0 & \dots & 0 \\ 0 & b_{22} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & b_{n n} \end{array}) \\ = (\begin{array}{c} a_{11} b_{11} & 0 & \dots & 0 \\ 0 & a_{22} b_{22} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & a_{n n} b_{n n} \end{array}) \end{aligned}

This completes the proof.

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Theorem.

Product of a triangular matrix and a diagonal matrix

Let $A$ be a triangular matrix and $D$ be a diagonal matrix:

if $A$ is upper triangular, then $A D$ and $D A$ are upper triangular matrices.
if $A$ is lower triangular, then $A D$ and $D A$ are lower triangular matrix.

Note that the diagonals of $A D$ and $D A$ are equal to the pairwise products of the diagonal entries of $A$ and $D$ .

Proof. Let $A$ be an upper triangular matrix and $D$ be a diagonal matrix. $D$ is also an upper triangular matrix by theoremlink, which means that the product $A D$ is upper triangular by theoremlink. We can also apply the same theoremlink to conclude that $D A$ is upper triangular. Finally, by theoremlink, the diagonal entries of $A D$ and $D A$ will be the pairwise products of the diagonal entries of $A$ and $D$ . The proof for the lower triangular case is analogous. This completes the proof.

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Theorem.

Inverse of a diagonal matrix

Let $D$ be a diagonal matrix:

D = (\begin{matrix} d_{11} & 0 & \dots & 0 \\ 0 & d_{22} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & d_{n n} \end{matrix})

If every diagonal entry of $D$ is non-zero, then $D^{- 1}$ is computed by:

D^{- 1} = (\begin{matrix} \frac{1}{d_{11}} & 0 & \dots & 0 \\ 0 & \frac{1}{d_{22}} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & \frac{1}{d_{n n}} \end{matrix})

This also means that the inverse of a diagonal matrix is also diagonal.

Proof. Suppose we have an $n \times n$ diagonal matrix $D$ and another matrix $A$ below:

\begin{array}{r} D = (\begin{array}{c} d_{11} & 0 & \dots & 0 \\ 0 & d_{22} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & d_{n n} \end{array}), A = (\begin{array}{c} \frac{1}{d_{11}} & 0 & \dots & 0 \\ 0 & \frac{1}{d_{22}} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & \frac{1}{d_{n n}} \end{array}) \end{array}

Our goal is to show that $A = D^{- 1}$ . The product $D A$ is:

\begin{array}{r} D A = (\begin{array}{c} d_{11} & 0 & \dots & 0 \\ 0 & d_{22} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & d_{n n} \end{array}) (\begin{array}{c} \frac{1}{d_{11}} & 0 & \dots & 0 \\ 0 & \frac{1}{d_{22}} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & \frac{1}{d_{n n}} \end{array}) \end{array}

By theoremlink, taking the product of two diagonal matrices involves multiplying the corresponding diagonal entries:

D A = (\begin{matrix} \frac{d_{11}}{d_{11}} & 0 & \dots & 0 \\ 0 & \frac{d_{22}}{d_{22}} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & \frac{d_{n n}}{d_{n n}} \end{matrix}) = (\begin{matrix} 1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & 1 \end{matrix}) = I_{n}

Because $D A = I_{n}$ , we have that $D^{- 1} = A$ 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:

D = (\begin{matrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{matrix})

Solution. The inverse of $D$ is a diagonal matrix whose diagonal entries are the reciprocal:

D^{- 1} = (\begin{matrix} 1 / 3 & 0 & 0 \\ 0 & 1 / 2 & 0 \\ 0 & 0 & 1 \end{matrix})

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Published by Isshin Inada

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