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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?

EDIT:

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.

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  • $\begingroup$ 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
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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?

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  • $\begingroup$ 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
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    $\begingroup$ @cdummie The eigenspace, yes. $\endgroup$ – amd Oct 7 '17 at 8:58
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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) $n-1$.

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  • $\begingroup$ 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

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