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.
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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.
\[\mathbf{x}^{\trans}A\mathbf{x}>0\] 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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[…] 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 […]