Inequality about Eigenvalue of a Real Symmetric Matrix

Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra

Problem 451

Let $A$ be an $n\times n$ real symmetric matrix.
Prove that there exists an eigenvalue $\lambda$ of $A$ such that for any vector $\mathbf{v}\in \R^n$, we have the inequality
\[\mathbf{v}\cdot A\mathbf{v} \leq \lambda \|\mathbf{v}\|^2.\]

 
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Proof.

Recall that all the eigenvalues of a symmetric matrices are real numbers.
Let $\lambda_1, \dots, \lambda_n$ be eigenvalues of $A$.

Since these eigenvalues are real numbers, there is the largest one.
Let $\lambda$ be the largest eigenvalue of $A$.
With this choice of $\lambda$ we show that the inequality
\[\mathbf{v}\cdot A\mathbf{v} \leq \lambda \|\mathbf{v}\|^2\] holds for any $\mathbf{v}\in \R^n$.


Also recall that for a real symmetric matrix, there are eigenvalues $\mathbf{v}_1, \dots, \mathbf{v}_n$ corresponding to $\lambda_1, \dots, \lambda_n$ such that
\[B=\{\mathbf{v}_1, \dots, \mathbf{v}_n\}\] form an orthonormal basis of $\R^n$.
(This statement is equivalent to that every real symmetric matrix is diagonalizable by an orthogonal matrix.)


Let $\mathbf{v}$ be an arbitrary vector in $\R^n$.
Then since $B$ is a basis of $\R^n$, we can write
\[\mathbf{v}=c_1\mathbf{v}_1+\dots+c_n\mathbf{v}_n\] for some $c_1, \dots, c_n\in \R$.

Then we calculate
\begin{align*}
A\mathbf{v}&=A(c_1\mathbf{v}_1+\dots+c_n\mathbf{v}_n)\\
&=c_1A\mathbf{v}_1+\dots+c_nA\mathbf{v}_n\\
&=c_1\lambda_1\mathbf{v}_1+\dots+c_n\lambda_n\mathbf{v}_n
\end{align*}
since $A\mathbf{v}_i=\lambda_i\mathbf{v}_i$ for $i=1, \dots, n$.


Using this, we have
\begin{align*}
\mathbf{v}\cdot A\mathbf{v}&=(c_1\mathbf{v}_1+\dots+c_n\mathbf{v}_n)\cdot (c_1\lambda_1\mathbf{v}_1+\dots+c_n\lambda_n\mathbf{v}_n)\\
&=c_1^2\lambda_1+\cdots+c_n^2\lambda_n.
\end{align*}

Here, we used that $B=\{\mathbf{v}_1, \dots, \mathbf{v}_n\}$ is an orthonormal basis of $\R^3$.
That is, we used the properties
\begin{align*}
\mathbf{v}_i\cdot \mathbf{v}_j=\begin{cases}
1 & \text{if } i=j\\
0 & \text{if } i\neq j.
\end{cases}
\end{align*}


Since $\lambda$ is the largest eigenvalue of $A$, we have
\begin{align*}
\mathbf{v}\cdot A\mathbf{v}&=c_1^2\lambda_1+\cdots+c_n^2\lambda_n\\
& \leq c_1^2\lambda+\cdots+c_n^2\lambda\\
&=\lambda(c_2^2+\cdots+c_n^2)\\
&=\lambda \|\mathbf{v}\|^2.
\end{align*}
Hence the required inequality holds.


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1 Response

  1. 06/12/2017

    […] For a proof of this problem, see the post “Inequality about Eigenvalue of a Real Symmetric Matrix“. […]

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