# Eigenvector matrix times diagonal matrix equals original matrix times eigenvector matrix?

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?

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

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).

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