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So I had a student come to me with a question which followed as such:

Given $\{c_1 , c_2 , c_3, \dots, c_n \}$ are real numbers, form the matrix A whose entries are given by $a_{ij} = c_i \cdot c_j $ from the above set of numbers (that is A is symmetric). The question asks that the student writes down every eigenvalue of the matrix explicitly and to basically reason out why there were that many eigenvalues, and why there cannot be more/less than that. Now this is a simpler undergraduate course in linear algebra, the knowledge the student has at their disposal is a limited chapter on eigenvalues and matrix diagonalization. I was able to help them understand how the determinant could be shown to be zero for this particular matrix, and through use of a pattern of ever growing $n \times n$ matrices guess that the characteristic polynomial of the matrix is given by $$\lambda^{n-1}(\lambda - Tr(A)) = 0 $$Which could be argued from the Caley-Hamilton Theorem. The idea is that the student doesn't know this and I am at a loss how to explain on a more basic undergrad level. Essentially all but the last eigenvalue are zero (if the matrix is not traceless).

Any help would be appreciated (I hope this is not a repeat, if so a pointer to the original question will also help)

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If you let $x=(c_1,c_2,\ldots,c_n)^T$, then $A=xx^T$. Note also that $x^T x = c_1^2+\cdots +c_n^2$.

Now $$ Ax=xx^Tx=(c_1^2+\cdots +c_n^2)\,x, $$ so $x$ is an eigenvector with eigenvalue $c_1^2+\cdots +c_n^2$.

If $y\perp x$, then $x^T y=0$, and so $Ay=xx^Ty=x0=0$.

If you form an orthogonal basis with first vector $x$, then each vector in the basis is an eigenvector for $A$, the first with eigenvalue $c_1^2+\cdots +c_n^2$ and the rest with eigenvalue $0$. Then, as you said, the characteristic polynomial is $$ \lambda^{n-1}(\lambda-\text{Tr}(A)). $$

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Hint: You can write the matrix $A$ as : $$ A=c^Tc. $$ where $c=[c_1 \dots c_n]$.

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  • $\begingroup$ Ah I don't know why I hadn't thought of that $\endgroup$ – Triatticus May 6 '14 at 2:28
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Here's a way to explain why there are $n-1$ linearly independent eigenvectors with eigenvalue $0$.

Notice that the $i$-th row is the vector $c_i ( c_1 , c_2 , \dots , c_n )$. Think of the vector $( c_1 , c_2 , \dots , c_n )$ resting in $n$-dimensional space. There is a "plane" of dimension $n-1$ that is orthogonal to this vector given by the equations $c_1 x_1 + c_2 x_2 + \dots + c_n x_n = 0$. So this is a vector subspace of dimension $n-1$, and has $n-1$ basis vectors which are linearly independent eigenvectors of our original matrix.

To find the last eigenvalue, use the eigenvector $( c_1 , c_2 , \dots c_n )$, which gives us an eigenvalue $\sum_{i=1}^n c_i^2$.

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