# Doubts about a question I asked a long time ago (eigenvalues)

Here I posted a question about the eigenvalues of the matrix $A:=vv^t$ (where $v\in\mathbb{R}^n$).

The question was answered but I think (after some time) that I am not satisfied.

Can someone please expand the answer? I don't understand why $A$ has rank at most $1$ and why this fact implies that $\lambda=\sum x_i^2$ is the unique eigenvalue. In addition, can I conclude that $A$ is diagonalizable?

• You mean $A=vv^T$ it's certainly not a dot product May 16 '18 at 19:55
• @NateIverson yes, this is not a dot product. Fixed! May 16 '18 at 19:57
• @N8tron if you want to be nitpicky about syntax, you should require it to be written $A = v\otimes v^t$. Anyways what other than that could $v\cdot v^t$ possibly be understood as? For a scalar product, it would need to be just $v\cdot v$. Some people would argue that this too should better be written $\langle v,v\rangle$. May 17 '18 at 10:21

Rank $1$: For any vector $a$, note that $v^Ta$ is a scalar, so $vv^Ta$ is a scalar multiple of $v$. So the range of $vv^T$ is within the span of $v$, which is one-dimensional, so $vv^T$ has rank $1$.

All but $1$ eigenvalue is $0$: Since the rank of $vv^T$ is $1$, the nullspace has dimension $n-1$, by the rank-nullity theorem. That means we can find $n-1$ linearly independent vectors in the nullspace. Since every vector in the nullspace has eigenvalue $0$, $0$ is an eigenvalue with multiplicity $n-1$. That leaves room for only one more eigenvalue, which we have already shown is $\sum x_i^2$.

Diagonalizability: As we have shown above, we can find $n-1$ linearly independent eigenvectors of $0$. We can also find one eigenvector of $\sum x_i^2$, to make a basis in which $vv^T$ is diagonal.

• Your answer help to me a lot. Thanks May 16 '18 at 20:09
• Shouldn't it be $\sum_iv_i^2$ instead of $\sum_ix_i^2$? I assume the magically appearing $x$ is actually $v$. The linked post says $v=(x_1,\dots,x_n)$ which resolves the issue, but it's a bit confusing for anyone who doesn't follow the link. (@Gödel The same goes for the question.) May 16 '18 at 21:19
• @JoonasIlmavirta $v_i$ vs. $x_i$ is just a matter of notational preference; neither is more correct than the other. Since the linked post and the question use $x_i$, I wouldn't want to suddenly change the notation to $v_i$. May 17 '18 at 0:30

A is diagonizable because it is real valued and symmetric $A^T=A$ (This is the spectral theorem)

$A=vv^T$ so $A^T=(vv^T)^T=(v^T)^Tv^T=vv^T=A$

https://en.wikipedia.org/wiki/Spectral_theorem

To see that $v v^T$ has rank at most one, we see that for any arbitrary vector $u$ ,we have: $$(vv^T)x = v(v^Tx)$$ The right hand side is a scalar times $v$. So every vector $x$ is mapped to a scalar multiple of $v$, with the scalar determined by $x$, meaning that the linear transformation represented by $vv^T$ has rank $1$ and so $vv^T$ has rank $1$.

Note that $$(vv^T)v = v(v^Tv) = \|v\|^2 v$$

so $v$ is an eigenvector with the eigenvalue $\|v\|^2 = \sum_{i=1}^n x_i^2$.

Also, explicitly

$$vv^T = \begin{pmatrix} x_1^2 & x_1x_2 & \ldots & x_1x_n \\ x_2x_1 & x_2^2 & \ldots & x_2x_n\\ \vdots & \vdots & \ddots & \vdots \\ x_nx_1 & x_nx_2 & \ldots & x_n^2\end{pmatrix}$$

so every column is a multiple of $v$, therefore the rank of $vv^T$ is at most $1$.

$A$ has rank at most $1$ because all rows are multiples of $(x_1,\dots,x_n)$.

Therefore, the kernel of $A$ has dimension at least $n-1$, leaving room for at most one other eigenvalue.

$A$ is diagonalizable by taking a basis of the kernel of $A$ and adding an eigenvector for $\lambda$.

You want to write $v$ as a column vector. Do the computation for any vector you like, and you will notice that the columns of $A$ are just a multiple of $v$.