# Positive definite Real Symmetric Matrix and its Eigenvalues

## Problem 396

A real symmetric $n \times n$ matrix $A$ is called 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.

## Proof.

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

Recall that the eigenvalues of a real symmetric matrix are real.
(See the corollary in the post “Eigenvalues of a Hermitian matrix are real numbers“.)

Let $\lambda$ be a (real) eigenvalue of $A$ and let $\mathbf{x}$ be a corresponding real eigenvector. That is, we have
$A\mathbf{x}=\lambda \mathbf{x}.$ Then we multiply by $\mathbf{x}^{\trans}$ on left and obtain
\begin{align*}
\mathbf{x}^{\trans}A\mathbf{x}&=\lambda \mathbf{x}^{\trans} \mathbf{x}\\
&=\lambda ||\mathbf{x}||^2.
\end{align*}

The left hand side is positive as $A$ is positive definite and $\mathbf{x}$ is a nonzero vector as it is an eigenvector.
Since the length $||\mathbf{x}||^2$ is positive, we must have $\lambda$ is positive.
It follows that every eigenvalue $\lambda$ of $A$ is real.

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

Note that a real symmetric matrix is diagonalizable by an orthogonal matrix.

So there exists an on orthogonal matrix $Q$ such that $Q^{\trans}AQ=D$, where $D$ is the diagonal matrix
$D=\begin{bmatrix} \lambda_1 & 0 & 0 & 0 \\ 0 &\lambda_2 & 0 & 0 \\ \vdots & \cdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{bmatrix}$ whose diagonal entries $\lambda_i$ are eigenvalues of $A$, which are positive by assumption.

Let $\mathbf{x}$ be an arbitrary nonzero vector in $\R^n$.
Since $A=QDQ^{\trans}$ (remark that $Q^{-1}=Q^{\trans}$), we have
$\mathbf{x}^{\trans} A\mathbf{x}=\mathbf{x}^{\trans}QDQ^{\trans}\mathbf{x}.$ Putting $\mathbf{y}=Q^{\trans}\mathbf{x}$, we can rewrite the above equation as
$\mathbf{x}^{\trans} A\mathbf{x}=\mathbf{y}^{\trans}D\mathbf{y}.$

Let
$\mathbf{y}=\begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix}.$ Then we have
\begin{align*}
&\mathbf{x}^{\trans} A\mathbf{x}=\mathbf{y}^{\trans}D\mathbf{y}\\
&=\begin{bmatrix}
y_1 & y_2 & \cdots & y_n
\end{bmatrix}
\begin{bmatrix}
\lambda_1 & 0 & 0 & 0 \\
0 &\lambda_2 & 0 & 0 \\
\vdots & \cdots & \ddots & \vdots \\
0 & 0 & \cdots & \lambda_n
\end{bmatrix}
\begin{bmatrix}
y_1 \\
y_2 \\
\vdots \\
y_n

### 2 Responses

1. 04/30/2017

[…] For a solution, see the post “Positive definite real symmetric matrix and its eigenvalues“. […]

2. 05/01/2017

[…] Recall that a symmetric matrix is positive-definite if and only if its eigenvalues are all positive. […]

This site uses Akismet to reduce spam. Learn how your comment data is processed.

##### If Two Vectors Satisfy $A\mathbf{x}=0$ then Find Another Solution

Suppose that the vectors \[\mathbf{v}_1=\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \qquad \mathbf{v}_2=\begin{bmatrix} -4 \\ 0...

Close