# Is it true that $V^{\,T}SV=\Lambda$?

Prompt

A symmetric matrix $$S=S^T$$ has orthonormal eigenvectors $$\vec{v}_1$$ to $$\vec{v}_n$$. Then any vector $$\vec{x}$$ can be written as a combination $$\vec{x} = c_1 \vec{v}_1+ \cdots + c_n \vec{x}_n$$. Explain this formula: $$\vec{x}^{\,T}S\vec{x} = \lambda_1 c_1^2+ \cdots + \lambda_nc_n^2$$

My explanation:

$$\vec{x}$$ can be written as $$\begin{bmatrix} \vec{v}_1 & \cdots & \vec{v}_n \end{bmatrix}_{nxn} \begin{bmatrix} c_1 \\ \vdots \\ c_n \end{bmatrix}_{\,nx1} = V\vec{c}$$ Consider, $$\vec{x}^{\,T}S\vec{x} = \vec{c}^{\,T}(V^{\,T}SV)\vec{c} = \vec{c}^{\,T}\Lambda\vec{c} = \lambda_1 c_1^2 + \cdots + \lambda c_n^2$$ We see that orthonormal eigenvectors of $$S$$ multiply to make the diagonal eigenvlaue matrix. We then end up with the inner product of the vector weightings, the $$c^2$$ terms, scaled by their respective eigenvalues.

I feel pretty good about my explanation. Just the part that I don't fully understand is why $$V^{\,T}SV=\Lambda$$... It seems to me like it should equal something more along the lines of $$\Lambda^T \Lambda$$ since each eigenvector should have been scaled when right multiplied because $$A\vec{v} = \lambda \vec{v}$$.

Is it true that $$V^{\,T}SV=\Lambda$$? If it is, a little help seeing why would be appreciated.

### A way I thought about it.

$$V^TSV = V^T(SV) = V^T \begin{bmatrix} \lambda_1 \vec{v_1} & ... & \lambda_n \vec{v_n} \end{bmatrix}$$ The way to think about $$SV$$ is that each column of $$V$$ is an eigenvector that multiplies the columns of $$S$$. We already know what this combination will equal by the definition of an eigenvector $$A\vec{v} = \lambda\vec{v}$$. Finally, when right multiplying by $$V^T$$ remember that $$\vec{v}_i^{\,T}\vec{v_j}=1$$ is $$1$$ when $$i=j$$ and $$0$$ when $$i\neq j$$ by definition of being orthonormal vectors.

Clearly we end up with $$\Lambda$$ as the result.

Yes, it is true that $$V^TSV = \Lambda$$. There are several ways that we can understand why this holds; one way is to compare what each matrix "does" to a column-vector.
Let $$\vec c$$ denote the column vector $$\vec c = (c_1,\dots,c_n)$$. Verify that $$\Lambda \vec c = (\lambda_1 c_1,\dots,\lambda_n c_n)$$.
Now, we consider the product $$V^TSV \vec c = V^T(S(V\vec c))$$. We find that $$V \vec c = \pmatrix{\vec v_1 & \cdots & \vec v_n} \pmatrix{c_1 \\ \vdots \\ c_n} = c_1 \vec v_1 + \cdots + c_n \vec v_n.$$ To put things another way, the role of the $$V$$ is to interpret the entries of $$\vec c$$ as the coefficients of the vectors $$\vec v_1,\dots,\vec v_n$$. From there, we see that because each $$v_i$$ is an eigenvector of $$S$$, we have $$S(V\vec c) = S(c_1 \vec v_1 + \cdots + c_n \vec v_n) = c_1 \lambda_1 \vec v_1 + \cdots + c_n \lambda_n \vec v_n.$$ Finally, note that (because $$V$$ is orthogonal) $$V^T$$ is the inverse of $$V$$. So, just as $$V$$ interprets the "input vector" as a list of coefficients for the $$v_i$$, $$V^T$$ takes a vector and gives us the the list of coefficients for the $$v_i$$ as its output. That is, $$V^T(S(V\vec c)) = (c_1 \lambda_1) \vec v_1 + \cdots + (c_n \lambda_n) \vec v_n = (c_1 \lambda_1,\dots,c_n \lambda_n).$$ So indeed, $$V^TSV$$ and $$\Lambda$$ describe the same transformation and are therefore the same matrix.
We can think of $$V^TSV$$ as an altered version of $$S$$ where, instead of thinking of the entries of a vector (both input and output) as the literal coordinates of a vector, we interpret them as the coefficients of our vectors $$v_i$$. In terms of terminology that you might already have heard before, we say that $$\Lambda$$ is the matrix of $$S$$ "after a change of basis".