Let $\hat{x} \in \mathbb{R^n}$ and $||\hat{x}||_2=1$. Determine kernel and image of matrix $A$ in terms of vector $\hat{x}$ if matrix $A$ is defined as $$A=\hat{x}\hat{x}^T \in \mathbb{R^n}$$ then find nullity and rank of the matrix.

After that find all eigenvalues of matrix $A$ and its algebraic and geometric multiplicities.

Basically, all i know here is the following:

Vector norm for which we know the value is Euclidian norm, so basically, $||\hat{x}||_2=1$ means $$\sqrt{x_1^2 + x_2^2 +...+x_n^2}=1$$ where $x_1, x_2,...,x_n$ are components of the given vector, now when we know this we know that matrix $A$ when defined as product of given vector and it's corresponding transposed vector is actually a orthogonal matrix whose trace is equal to one. However, i don't know how can i use this to find kernel and image in terms of given vector and nullity, rank and eigenvalues and their algebraic and geometric multiplicities for this matrix. How can i solve this?


Ok, so $$A=\begin{bmatrix} x_1^2 & x_1x_2 &...&x_1x_n\\x_2x_1 &x_2^2 &... & x_2x_n \\.\\.\\.\\ x_nx_1 &x_nx_2 &...&x_n^2\end{bmatrix}$$

Which means that this is the matrix for which i should determine kernel, image and the rest of the things noted above, but even now, i don't understand how.

  • $\begingroup$ Do you know what this matrix would represent? It's the projection of a vector onto $\hat{x}$. What do you know about the kernel and image of this? If you didn't know what this matrix represents, note that the columns of $\hat{x} \hat{x}^T$ are just scalar multiples of $\hat{x}$. So the rank is $1$ and the nullity is $n-1$. So the image is spanned by $\hat{x}$. See if you can figure out what the kernel is! $\endgroup$ – Osama Ghani Oct 4 '17 at 15:36
  • $\begingroup$ @OsamaGhani I've edited my post because my answer wouldn't fit here on comments section $\endgroup$ – cdummie Oct 4 '17 at 17:34

Observe that every column of the matrix is a scalar multiple of $x$, so its column space is simply the span of $x$. We know that $x\ne0$, so the rank of $A$ is obviously one, with nullity $n-1$. This makes zero an eigenvalue with associated eigenspace the null space of $A$. The sum of the eigenvalues of a matrix is equal to its trace, so the remaining eigenvalue is $\sum_{i=0}^nx_i^2=\|x\|_2^2=1$. Can you find an associated eigenvector?

  • $\begingroup$ Lets see, if we say that $x_1, x_2, ..., x_n$ are components of vector $x$ then columns of matrix $A$ are actually $x_1*x, x_2*x, ... x_n*x$ and it makes them linearly dependent, so that's why $rankA=1$ , since A is nxn matrix, nullity is n-1 and since nullity of A is dimension of following set $x: Ax=0$ it means that zero is eigenvalue with alg. multiplicity n-1. Now, for eigenvalue 1 we have $Ax=x$ so eigenvector i am looking for should be solution to this equation, i am just not sure how to find it. $\endgroup$ – cdummie Oct 6 '17 at 16:25
  • $\begingroup$ @cdummie Hint: $Av=\hat x\hat x^T\hat v=(\hat x^T\hat v)\hat x$. $\endgroup$ – amd Oct 6 '17 at 19:19
  • $\begingroup$ Are you suggesting that since we want to find $Av=\hat{x}$ then $\hat{x}\hat{x}^Tv=\hat{x}$ so this holds for every $v$ such that $\hat{x}^Tv=1$? $\endgroup$ – cdummie Oct 7 '17 at 6:38
  • $\begingroup$ @cdummie Not at all. $A\hat v$ is always a multiple of $\hat x$, so the only possibility for an eigenvector of $1$ is some multiple of $\hat x$. $\endgroup$ – amd Oct 7 '17 at 7:18
  • 1
    $\begingroup$ @cdummie The eigenspace, yes. $\endgroup$ – amd Oct 7 '17 at 8:58

Let $y\in\ker A$, that is, let $y$ be a vector such that $(\hat x.\hat x^T).y=0$. But $(\hat x.\hat x^T).y=\hat x.(\hat x^T.y)$, and therefore any vector orthogonal to $\hat x$ belongs to $\ker A$. Since $\dim\{\hat x\}^\perp=n-1$, $\dim\ker f=n-1$ or $\dim\ker A=n$. But if $\dim\ker A=n$, then $A$ would be the null function, which is not, since $A.x=x$. Therefore, $\dim\ker f=n-1$ and $\ker f=\{\hat x\}^\perp$.

Note that, as I wrote, $A.x=x$. So, $1$ is an eigenvalue, with multiplicity (algebraic and geometric) $1$, and the only other eigenvalue is $0$, with multiplicity (again, algebraic and geometric) $n-1$.

  • $\begingroup$ But i still don't understand how dimension of kernel is n-1, and what do you mean by $dim{\hat{x}}$. Is that the rank of it or what? $\endgroup$ – cdummie Oct 4 '17 at 17:38
  • $\begingroup$ @cdummie What is the passage that you don't understand in my proof of the fact that $\dim\ker A=n-1$? Besides, at no point I wrote $\dim\hat x$. $\endgroup$ – José Carlos Santos Oct 4 '17 at 17:44
  • $\begingroup$ Well, you said, since $dim[ \hat{x} ]=n-1$, $dimkerf=n-1$ or $dimkerA=n$. To be honest, i don't understand any of it, i mean i understand the part when you say that any vector orthogonal to $\hat{x}$ belongs to $kerA$, that's fine, but i don't understand the rest, and what is the difference between $dim\hat{x}$ and $dim[ \hat{x} ]$ $\endgroup$ – cdummie Oct 4 '17 at 18:09
  • $\begingroup$ @cdummie What I wrote was that $\dim\{\hat x\}^\perp=n-1$. Since $\{\hat x\}$ is a set with a single non-null vector, I don't even know what $\dim\{\hat x\}$ means. Besides, $\{\hat x\}^\perp$ is the set of all vectors orthogonal to $\hat x$, and it happens to be a vector space, $\endgroup$ – José Carlos Santos Oct 4 '17 at 18:13
  • $\begingroup$ Ok, so how you found out that $dim{\hat}\bot=n-1$? I suppose that it is dimension of orthogonal complement of $\hat{x}$ $\endgroup$ – cdummie Oct 4 '17 at 18:17

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.