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?
Best Answer
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).