# 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.

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## 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

\end{bmatrix}\\[6pt]
&=\lambda_1y_1^2+\lambda_2 y_2^2+\cdots +\lambda_n y_n^2.

\end{align*}

By assumption eigenvalues $\lambda_i$ are positive.

Also, since $\mathbf{x}$ is a nonzero vector and $Q$ is invertible, $\mathbf{y}=Q^{\trans}\mathbf{x}$ is not a zero vector.

Thus the sum expression above is positive, hence $\mathbf{x}^{\trans} A\mathbf{x}$ is positive for any nonzero vector $\mathbf{x}$.

Therefore, the matrix $A$ is positive-definite.

## Related Questions.

**Problem**. 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.

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

**Problem**. Suppose $A$ is a positive definite symmetric $n\times n$ matrix.

**(a)** Prove that $A$ is invertible.

**(b)** Prove that $A^{-1}$ is symmetric.

**(c)** Prove that $A^{-1}$ is positive-definite.

For proofs, see the post “Inverse matrix of positive-definite symmetric matrix is positive-definite“.

**Problem**.

Prove that a positive definite matrix has a unique positive definite square root.

For a solution of this problem, see the post

A Positive Definite Matrix Has a Unique Positive Definite Square Root

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