# 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
\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 ### More from my site • Inverse Matrix of Positive-Definite Symmetric Matrix is Positive-Definite 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. (MIT, Linear Algebra Exam Problem) Proof. (a) Prove that A is […] • Transpose of a Matrix and Eigenvalues and Related Questions 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 […] • Eigenvalues of a Hermitian Matrix are Real Numbers Show that eigenvalues of a Hermitian matrix A are real numbers. (The Ohio State University Linear Algebra Exam Problem) We give two proofs. These two proofs are essentially the same. The second proof is a bit simpler and concise compared to the first one. […] • Eigenvalues of 2\times 2 Symmetric Matrices are Real by Considering Characteristic Polynomials Let A be a 2\times 2 real symmetric matrix. Prove that all the eigenvalues of A are real numbers by considering the characteristic polynomial of A. Proof. Let A=\begin{bmatrix} a& b \\ c& d \end{bmatrix}. Then […] • A Symmetric Positive Definite Matrix and An Inner Product on a Vector Space (a) Suppose that A is an n\times n real symmetric positive definite matrix. Prove that \[\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y} defines an inner product on the vector space $\R^n$. (b) Let $A$ be an $n\times n$ real matrix. Suppose […]
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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. […]

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