# Transpose of a Matrix and Eigenvalues and Related Questions

## Problem 12

Let $A$ be an $n \times n$ real matrix. Prove the followings.

**(a)** The matrix $AA^{\trans}$ is a symmetric matrix.

**(b) **The set of eigenvalues of $A$ and the set of eigenvalues of $A^{\trans}$ are equal.

**(c)** The matrix $AA^{\trans}$ is non-negative definite.

(An $n\times n$ matrix $B$ is called *non-negative definite* if for any $n$ dimensional vector $\mathbf{x}$, we have $\mathbf{x}^{\trans}B \mathbf{x} \geq 0$.)

**(d)** All the eigenvalues of $AA^{\trans}$ is non-negative.

Contents

## Facts.

Before the proofs, we first review several basic properties of the transpose of a matrix.

- For any matrices $A$ and $B$ so that the product $AB$ is defined, we have $(AB)^{\trans}=B^{\trans}A^{\trans}$
- We have $(A^{\trans})^{\trans}=A$ for any matrix $A$.

Also recall that the eigenvalues of a matrix $A$ are the solutions of the characteristic polynomial $p_A(x)=\det(A-xI)$.

## Proof.

### (a) The matrix $AA^{\trans}$ is a symmetric matrix.

We compute $(AA^{\trans})^{\trans}=(A^{\trans})^{\trans}A^{\trans}=AA^{\trans}$ and thus $AA^{\trans}$ is a symmetric matrix.

(We used the Fact 1 and 2.)

### (b) The set of eigenvalues of $A$ and the set of eigenvalues of $A^{\trans}$ are equal.

We show that the characteristic polynomials of $A$ and $A^{\trans}$ are the same, hence they have exactly same eigenvalues.

Let $p_A(x)$ and $p_{A^{\trans}}(x)$ be the characteristic polynomials of $A$ and $A^{\trans}$, respectively. Then we have

\begin{align*}

p_A(x)=\det(A-xI)=\det(A-xI)^{\trans} =\det(A^{\trans}-xI)=p_{A^{\trans}}(x).

\end{align*}

The first and last equalities are the definition of the characteristic polynomial.

The second equality is true because in general we have $\det(B)=\det(B^{\trans})$ for a square matrix $B$. This completes the proof of (b).

### (c) The matrix $AA^{\trans}$ is non-negative definite.

Let $\mathbf{x}$ be an $n$ dimensional vector. Then we have

\begin{align*}

\mathbf{x}^{\trans}AA^{\trans}\mathbf{x}=(A^{\trans}\mathbf{x})^{\trans}(A^{\trans}\mathbf{x})=||A^{\trans}\mathbf{x}|| \geq 0,

\end{align*}

since a norm (length) of a vector is always non-negative.

Thus $AA^{\trans}$ is non-negative definite.

### (d) All the eigenvalues of $AA^{\trans}$ is non-negative.

Let $\lambda$ be an eigenvalue of $AA^{\trans}$ and let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$.

Then we compute

\begin{align*}

\mathbf{x}^{\trans}AA^{\trans}\mathbf{x}=\mathbf{x}^{\trans}\lambda\mathbf{x}=\lambda ||\mathbf{x}||.

\end{align*}

Here the first equality follows from the definitions of the eigenvalue $\lambda$ and eigenvector $\mathbf{x}$. In part (c), we proved that $AA^{\trans}$ is non-negative definite, hence we have $\lambda ||\mathbf{x}|| \geq 0$.

Therefore $\lambda \geq 0$ and this completes the proof of (d).

## Related Question.

**positive definite**if

\[\mathbf{x}^{\trans}A\mathbf{x}>0\] for all nonzero vectors $\mathbf{x}$ in $\R^n$.

**(a)** Prove that the eigenvalues of a real symmetric positive-definite matrix $A$ are all positive.

**(b)** Prove that if eigenvalues of a real symmetric matrix $A$ are all positive, then $A$ is positive-definite.

For a solution, see the post ↴

“Positive definite real symmetric matrix and its eigenvalues“.

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## 2 Responses

[…] For a solution, see the post “Transpose of a matrix and eigenvalues and related questions.“. […]

[…] eigenvalues , we deduce that the matrix $A$ has an eigenvalue $1$. (See part (b) of the post “Transpose of a matrix and eigenvalues and related questions.“.) Let $mathbf{x}$ be an eigenvector corresponding to the eigenvalue $1$ (by definition […]