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How can I find the characteristic polynomial of the matrix $A$, so that I can find all of its eigenvalues, and hence the spectrum, so that I can use the spectrum to calculate the determinant of $A$?

And $A$ is a $10 \times 10$ matrix with $2$'s on the main diagonal and $1$'s everywhere else.

user321401
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2 Answers2

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Recall that for a square matrix $A$, eigenvectors and eigenvalues satisfy the equation

$$Ax = \lambda x.$$

Moving everything to one side, you get

$$Ax-\lambda x = 0.$$

Factoring out the $x$, you have

$$(A-\lambda I)x = 0.$$

To solve for $x$, if $A-\lambda I$ were an invertible matrix, then multiplication by its inverse gives

$$x = (A-\lambda I)^{-1}0 = 0$$

which is not particularly interesting. So we instead choose $\lambda$ so that the matrix $A-\lambda I$ is not invertible. So eigenvalues are precisely the $\lambda$ that make the matrix $A-\lambda I$ not invertible and the determinant gives a nice criterion for when this is true.

Taking $A$ to be the matrix in the question, it is clear that $A-I$ is the matrix of all $1$'s. Every row is a linear combination of each other so the matrix is not invertible and $\lambda = 1$ is an eigenvalue. This is obvious by simply looking at the matrix $A$ and asking ourselves if there is a constant we can subtract along the main diagonal so that two rows are the same.

It is clear that there is only one linearly independent row. So the dimension of the eigenspace is $9$. Now use the fact that

$$\operatorname{tr}(A) = \sum_{\lambda \in \sigma (A)} \lambda$$

and the knowledge of the other nine eigenvalues to get the tenth.

JessicaK
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  • Okay, I understand now that $A-I$ consists of rows that are linearly dependent (since all the rows of $A-I$ are the same), and this prompts us to see that $\lambda=1$ must be one of the eigenvalues of $A$. Now, the trace of the matrix of $A$ is $20$, and that is the sum of the eigenvalues of $A$. At this point, I should be able to find the rest of the nine remaining eigenvalues. Is my understanding correct so far? – user321401 Mar 15 '16 at 08:27
  • @user321401 You already know 9 of the eigenvalues are $1$ because if you take $A-I$ to reduced row echelon form, there are 9 rows of $0$'s. This is easily done in your head since all the rows are the same. – JessicaK Mar 15 '16 at 08:30
  • Okay, so 9 of the eigenvalues are $1$. Since the trace of $A$ in my case is equal to $20$, the last eigenvalue must be $11$. So my spectrum is ${1,1,1,1,1,1,1,1,1,11}$, right? – user321401 Mar 15 '16 at 09:08
  • @user321401 Indeed. – JessicaK Mar 15 '16 at 09:14
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You can write your matrix as

$$A=I+a\otimes a;\quad a=(1,1,1,1,1,1,1,1,1,1)$$

(identity + projection to vector $a$, you can also write by components, $A_{ij}=\delta_{ij}+a_i a_j$).

From symmetry argument, $a$ has a special role (1D eigenspace), and all the vectors, perpendicular to $a$ will behave equivalently (and thus have the same eigenvalue). Then just try what happens for both cases:

$$Aa=a+|a|^2 a = \underbrace{(1+|a|^2)}_{\lambda=11}a$$ $$Aa_\perp=a+\underbrace{(a\cdot a_\perp)}_0 a=\underbrace{1}_{\lambda=1}a$$ So eigenvalues are (11,1,1,1,1,1,1,1,1,1). You can get the characteristic polynomial yourself.

orion
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  • The characteristic polynomial is $(x-1)^9(x-11)$, and the determinant of $A$ is $11$? – user321401 Mar 15 '16 at 08:21
  • Precisely. And you also got all the eigenvectors for free (a for λ=11 and orthogonal complement of a for λ=1). – orion Mar 15 '16 at 08:23