3
$\begingroup$

Suppose we have a $n\times n$ real symmetric matrix $A$, and $V=[v_1\ v_2\cdots v_n]$ whose columns are the eigenvectors corresponding to the $n$ eigenvalues $\lambda_1,\ldots,\lambda_n$. Let $D$ be a diagonal matrix ${\rm diag}[\lambda_1\cdots\lambda_n]$, where $\lambda_i$ are the eigenvalues of $A$ for $i=1,\ldots,n$.

How do we prove that $A𝑉 = 𝑉D$?

edit: I realized what I said below is incorrect because multiplying $𝑉D$ does not give a diagonal matrix. But I am still confused as how I would know that $A𝑉 = 𝑉D$ when not given any numbers.

I was trying to start this proof with expanding $𝑉D$ and I got another diagonal matrix with the diagonal entries being $v_1\lambda_1,\ldots,v_n\lambda_n$ and all other entries being $0$. I am unsure how this could be equivalent to A𝑉 because wouldn't this resultant matrix not be a diagonal matrix? Or how would we know without knowing what the entries of A are?

$\endgroup$
2
  • $\begingroup$ What do you mean by vectors ($v_j\lambda_j$) on the diagonal? $\endgroup$
    – Graf Zahl
    Sep 20 at 22:32
  • $\begingroup$ Sorry, I realized what I said was wrong, I edited my post now! $\endgroup$
    – user774633
    Sep 21 at 0:11
1
$\begingroup$

Let $A$ be any $n \times n$ matrix, symmetric or not, over any field $\Bbb F$, and suppose that $A$ is possessed of $n$ eigenvectors $v_1$, $v_2$, $\ldots$, $v_n$ with the corresponding eigenvalues $\lambda_1$, $\lambda_2$, $\ldots$, $\lambda_n$ such that

$Av_i = \lambda_i v_i, \; 1 \le i \le n; \tag 1$

with our OP user 774633 we define the matrix whose columns are the eigenvectors $v_i$:

$V = [v_1 \; v_2 \; \ldots \; v_n], \tag 2$

and observe that

$AV = [Av_1 \; Av_2 \; \ldots \; Av_n]; \tag 3$

now in accord with (1) we may write

$AV = [\lambda_1 v_1 \; \lambda_2 v_2 \; \ldots \; \lambda_n v_n]. \tag 4$

Now again in accord with our OP we set

$D = \text{diag}[\lambda_1 \; \lambda_2 \; \ldots\; \lambda_n], \tag 5$

and we have

$VD = [v_1 \; v_2 \; \ldots \; v_n]\text{diag}[\lambda_1 \; \lambda_2 \; \ldots\; \lambda_n], \tag 6$

and if we write $V$ and $D$ in full matrix form (which explicitly presents every element) we obtain

$V = \begin{bmatrix} v_{11} & v_{12} & \ldots & v_{1n} \\ v_{21} & v_{22} & \ldots & v_{2n} \\ \vdots & \vdots & \ldots & \vdots \\ v_{n1} & v_{n2} & \ldots & v_{nn} \end{bmatrix} = [v_{ij}], \tag 7$

and

$D = \begin{bmatrix} \lambda_1 & 0 & 0 & \ldots & 0 \\ 0 & \lambda_2 & 0 & \ldots & 0 \\ 0 & 0 & \lambda_3 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ldots & \vdots \\ 0 & 0 & 0 & \ldots & \lambda_n \end{bmatrix} = [\delta_{ij} \lambda_i]. \tag 8$

Note that the row indices in the matrix (7) are the first or $i$ indices in the entries $v_{ij}$, in conformance with the standard practice for writing out matrices; adopting this convention clarifies the ensuing calculations.

We may thus exploit the ordinary rule for matrix multiplication and we find

$VD = [v_{ij}][\delta_{ij} \lambda_i] = \left [ \displaystyle \sum_{k = 1}^n v_{ik}\delta_{kj}\lambda_k \right] = [v_{ij} \lambda_j], \tag 9$

and it is clear that

$[v_{ij} \lambda_j] = [\lambda_1 v_1 \; \lambda_2 v_2 \; \ldots \; \lambda_n v_n] , \tag{10}$

whence, taking (3),(4) and (9) in concert, we arrive at

$AV = VD, \tag{11}$

the requisite relation 'twixt $A$, $V$, and $D$.

Perhaps a somewhat more elegant demonstration of (11) may be had via the observation that the $j$-th column of the matrix $D$ is in fact the vector

$\mathbf e_j = [\delta_{ij}], \tag{12}$

comprised of all $0$s save for a single $1$ in the $j$-th row, multiplied by the scalar $\lambda_j$, that is, $\lambda_j \mathbf e_j$. We may thus write

$D = \begin{bmatrix}\lambda_1 \mathbf e_1 & \lambda_2 \mathbf e_2 & \ldots & \lambda_n \mathbf e_n \end{bmatrix}; \tag{13}$

it is furthermore easy to see that

$V\mathbf e_i = v_i, \tag{14}$

whence

$VD = V\begin{bmatrix}\lambda_1 \mathbf e_1 & \lambda_2 \mathbf e_2 & \ldots & \lambda_n \mathbf e_n \end{bmatrix} = \begin{bmatrix}\lambda_1 V\mathbf e_1 & \lambda_2 V\mathbf e_2 & \ldots & \lambda_n V\mathbf e_n \end{bmatrix}$ $= \begin{bmatrix}\lambda_1 v_1 & \lambda_2 v_2 & \ldots & \lambda_n v_n \end{bmatrix} = \begin{bmatrix} Av_1 & Av_2 & \ldots & Av_n \end{bmatrix} = AV \tag{15}$

in accord with (3) and (4).

$\endgroup$
1
$\begingroup$

You can check equality of the columns by multiplying the matrices with the canonical basis vectors $e_i$. Then $AVe_i=Av_i=\lambda_iv_i = V (\lambda_i e_i)=V D e_i$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.