Matrix $\boldsymbol{A}$ is called symmetric if and only if $\boldsymbol{A}^T=\boldsymbol{A}$. This means that a matrix is symmetric if the matrix remains unchanged after flipping the rows and columns.

Example.

Symmetric matrices

The following are examples of symmetric matrices:

$$\begin{pmatrix}1&3\\3&2\end{pmatrix} ,\;\;\;\;\;\;\; \begin{pmatrix} 1&2&3\\2&5&6\\3&6&9 \end{pmatrix},\;\;\;\;\;\;\; \begin{pmatrix} 2&0&0\\0&3&0\\0&0&0 \end{pmatrix}$$

Theorem.

Symmetric matrices are square

If $\boldsymbol{A}$ is a symmetric matrix, then $\boldsymbol{A}$ is square.

Proof. Let $\boldsymbol{A}$ be an $m\times{n}$ matrix, which means that the shape of $\boldsymbol{A}^T$ is $n\times{m}$. By definition, if matrix $\boldsymbol{A}$ is symmetric, then $\boldsymbol{A}^T=\boldsymbol{A}$. For this equality to hold, the shapes of $\boldsymbol{A}^T$ and $\boldsymbol{A}$ must match, that is, $m=n$. This completes the proof.

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Theorem.

Sum of two symmetric matrices is symmetric

If $\boldsymbol{A}$ and $\boldsymbol{B}$ are symmetric matrices, then $\boldsymbol{A}+\boldsymbol{B}$ is also symmetric.

Proof. Let $\boldsymbol{A}$ and $\boldsymbol{B}$ be symmetric matrices. To show that $\boldsymbol{A}+\boldsymbol{B}$ is symmetric, we must show that the following holds:

$$(\boldsymbol{A}+\boldsymbol{B})^T =\boldsymbol{A}+\boldsymbol{B} $$

By the propertylink of transpose, we have that:

$$\begin{align*} (\boldsymbol{A}+\boldsymbol{B})^T &=\boldsymbol{A}^T+\boldsymbol{B}^T \end{align*}$$

Because $\boldsymbol{A}$ and $\boldsymbol{B}$ are symmetric, $\boldsymbol{A}^T=\boldsymbol{A}$ and $\boldsymbol{B}^T=\boldsymbol{B}$. Therefore, we end up with:

$$\begin{align*} (\boldsymbol{A}+\boldsymbol{B})^T &=\boldsymbol{A}+\boldsymbol{B} \end{align*}$$

This completes the proof.

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Theorem.

Scalar multiples of a symmetric matrix is also symmetric

If $\boldsymbol{A}$ is a symmetric matrix, then $k\boldsymbol{A}$ is also symmetric for any scalar $k$.

Proof. Let $\boldsymbol{A}$ be a symmetric matrix and $k$ be any scalar. By the propertylink of matrices, we have that:

$$\begin{align*} (k\boldsymbol{A})^T&= k\boldsymbol{A}^T\\ &=k\boldsymbol{A}\\ \end{align*}$$

By definition of symmetric matrices, $k\boldsymbol{A}$ is symmetric. This completes the proof.

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Theorem.

If A and B are symmetric, then AB+BA is symmetric

If $\boldsymbol{A}$ and $\boldsymbol{B}$ are symmetric matrices, then $\boldsymbol{AB}+\boldsymbol{BA}$ is a symmetric matrix.

Proof. The transpose of $\boldsymbol{AB}+\boldsymbol{BA}$ is:

$$\begin{align*} (\boldsymbol{AB}+\boldsymbol{BA})^T &=(\boldsymbol{AB})^T+(\boldsymbol{BA})^T\\ &=\boldsymbol{B}^T\boldsymbol{A}^T+ \boldsymbol{A}^T\boldsymbol{B}^T\\ &=\boldsymbol{B}\boldsymbol{A}+ \boldsymbol{A}\boldsymbol{B}\\ &=\boldsymbol{A}\boldsymbol{B}+ \boldsymbol{B}\boldsymbol{A} \end{align*}$$

Here:

the first equality follows from theoremlink.
the second equality follows from theoremlink.

By definition of symmetric matrices, we conclude that $\boldsymbol{AB}+\boldsymbol{BA}$ is symmetric. This completes the proof.

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Theorem.

Powers of symmetric matrices is also symmetric

If $\boldsymbol{A}$ is a symmetric matrix, then $\boldsymbol{A}^n$ is symmetric where $n$ is any positive integer.

Proof. The transpose of $\boldsymbol{A}^n$ is:

$$\begin{align*} (\boldsymbol{A}^n)^T &=(\boldsymbol{A}\boldsymbol{A}\cdots\boldsymbol{A})^T\\ &=\boldsymbol{A}^T\boldsymbol{A}^T\cdots\boldsymbol{A}^T\\ &=\boldsymbol{A}\boldsymbol{A}\cdots\boldsymbol{A}\\ &=\boldsymbol{A}^n \end{align*}$$

Here, we used theoremlink for the second equality. $\boldsymbol{A}^n$ is thus symmetric by definition. This completes the proof.

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Theorem.

Symmetric matrix is invertible if and only if its inverse is symmetric

Let $\boldsymbol{A}$ be a symmetric matrix. $\boldsymbol{A}$ is invertible if and only if $\boldsymbol{A}^{-1}$ is symmetric.

Proof. By propertylink of invertible matrices:

$$\begin{align*} (\boldsymbol{A}^{-1})^{T}&= (\boldsymbol{A}^{T})^{-1}\\ &= \boldsymbol{A}^{-1} \end{align*}$$

By definition, $\boldsymbol{A}^{-1}$ is therefore symmetric. The backward proposition also holds because every step is based on equality. This completes the proof.

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Theorem.

Product of symmetric matrices is commutative if and only if the product is symmetric

Let $\boldsymbol{A}$ and $\boldsymbol{B}$ be two symmetric matrices. $\boldsymbol{AB}=\boldsymbol{BA}$ if and only if $\boldsymbol{AB}$ is symmetric.

Proof. We will first prove the forward proposition. Assume $\boldsymbol{AB}=\boldsymbol{BA}$. The transpose of $\boldsymbol{AB}$ is:

$$\begin{align*} (\boldsymbol{AB})^T&= \boldsymbol{B}^T\boldsymbol{A}^T\\ &=\boldsymbol{B}\boldsymbol{A}\\ &=\boldsymbol{A}\boldsymbol{B} \end{align*}$$

By definition, $\boldsymbol{AB}$ is therefore symmetric.

Next we will prove the converse. Assume $\boldsymbol{AB}$ is symmetric. $\boldsymbol{AB}$ is:

$$\begin{align*} \boldsymbol{BA} &=\boldsymbol{B}^T\boldsymbol{A}^T\\ &=(\boldsymbol{A}\boldsymbol{B})^T\\ &=\boldsymbol{A}\boldsymbol{B} \end{align*}$$

This completes the proof.

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Theorem.

Symmetric matrices and dot products

An $n\times{n}$ matrix $\boldsymbol{A}$ is symmetric if and only if the following is true:

$$(\boldsymbol{Ax})\cdot \boldsymbol{y}= \boldsymbol{x}\cdot(\boldsymbol{Ay})$$

Where $\boldsymbol{x}$ and $\boldsymbol{y}$ are any vectors in $\mathbb{R}^n$.

Proof. We prove the forward proposition first. Suppose $\boldsymbol{A}$ is an $n\times{n}$ symmetric matrix. For any vectors $\boldsymbol{x},\boldsymbol{y}\in\mathbb{R}^n$, we have that:

$$\begin{align*} (\boldsymbol{Ax})\cdot{\boldsymbol{y}} &=(\boldsymbol{Ax})^T{\boldsymbol{y}}\\ &=\boldsymbol{x}^T\boldsymbol{A}^T{\boldsymbol{y}}\\ &=\boldsymbol{x}^T\boldsymbol{A}{\boldsymbol{y}}\\ &=\boldsymbol{x}\cdot(\boldsymbol{A}{\boldsymbol{y}})\\ \end{align*}$$

Note that the final step follows by theoremlink.

Now, let's prove the converse:

$$\begin{align*} (\boldsymbol{Ax})\cdot \boldsymbol{y}&= \boldsymbol{x}\cdot(\boldsymbol{Ay})\\ \boldsymbol{y}^T\boldsymbol{Ax} &=(\boldsymbol{Ay})^T\boldsymbol{x}\\ \boldsymbol{y}^T\boldsymbol{Ax}&= \boldsymbol{y}^T\boldsymbol{A}^T\boldsymbol{x}\\ \boldsymbol{y}^T\boldsymbol{Ax}- \boldsymbol{y}^T\boldsymbol{A}^T\boldsymbol{x}&=\boldsymbol{0}\\ \boldsymbol{y}^T(\boldsymbol{A}-\boldsymbol{A}^T)\boldsymbol{x}&=\boldsymbol{0} \end{align*}$$

Therefore $\boldsymbol{A}=\boldsymbol{A}^T$, which means $\boldsymbol{A}$ is symmetric. This completes the proof.

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Theorem.

Product of a matrix and its transpose is symmetric

If $\boldsymbol{A}$ is an $m\times{n}$ matrix, then $\boldsymbol{A}^T\boldsymbol{A}$ and $\boldsymbol{AA}^T$ are symmetric.

Proof. If $\boldsymbol{A}$ is an $m\times{n}$ matrix, then the shape of $\boldsymbol{A}^T$ is $n\times{m}$. Therefore, the shape of $\boldsymbol{A}^T\boldsymbol{A}$ is $m\times{m}$, which means $\boldsymbol{A}^T\boldsymbol{A}$ is square. Next, $\boldsymbol{A}^T\boldsymbol{A}$ can be written as:

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

Here, the first equality holds by propertylink of transpose. Since the transpose of $\boldsymbol{A}^T\boldsymbol{A}$ is itself, $\boldsymbol{A}^T\boldsymbol{A}$ is symmetric by definitionlink. Similarly, we can easily show that $\boldsymbol{AA}^T$ is symmetric:

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

By definitionlink, $\boldsymbol{AA}^T$ is symmetric. This completes the proof.

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Published by Isshin Inada

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