# linear algebra characteristic polynomial, matrix rank, Matrix similarity

I'm given 2 $n \times n$ matrices, $n>1$.

A=$\begin{bmatrix}1 & .& .& .& .& .& 1\\. &&&&&&.\\.&&&&&&.\\.&&&&&&.\\.&&&&&&.\\.&&&&&&.\\1 &.&.&.&.&.&1\end{bmatrix}$

B=$\begin{bmatrix}n & 0& & .& .& 0& 0\\0 &0&&&&&.\\.&&&&&&.\\.&&&&&&.\\.&&&&&&.\\.&&&&&&.\\0 &.&.&.&.&.&0\end{bmatrix}$

1) I need to find the characteristic polynomial of A using A's Rank.

2) I need to prove that the Coefficient of $t^n-1$ in the characteristic polynomial of A is equal -(trA).

3) I need to prove that A and B are similar matrices and find P so that $B = P^{-1}AP$

*All of A's entries = 1.

• Are all of the entries of $A$ equal to $1$ or just the ones on the boundary? – JimmyK4542 Sep 10 '14 at 18:55
• Question 2 asks to prove something which is false. The characteristic polynomial is incorrect and it certainly can't equal the constant $-\text{tr}(A)$. – Git Gud Sep 10 '14 at 19:01
• I fixed question 2 it asks fo the coefficient mabey now it makes more sense – durian Sep 10 '14 at 19:07
• I think maybe question 2 should be "coefficient of $t^{n-1}$"? – BaronVT Sep 10 '14 at 19:08

$A$ is symmetric, so the algebraic multiplicity of an eigenvalue is equal to the geometric multiplicity.

It is not hard to see that, for any $x$, $Ax = c(1,1,\dots,1)^T$ for some constant $c$. Thus, its rank is $1$ (corresponding to eigenvalue $\lambda = ...?$) and the other $n-1$ eigenvalues are $0$. Such a matrix has characteristic polynomial

$$(t - \lambda)(t - 0)^{n-1} = t^{n-1}(t - \lambda)$$

For question 2, it is easy to directly calculate the trace, and you should now have the characteristic polynomial, so just verify.

To find a similarity transform, you can find all the eigenvectors (meaning, find $n$ linearly independent eigenvectors) of $A$, or of $B$. One will be much easier than the other.

• Can you please explain why 1 is the rank of lambda? I know that $dimVλ = dimP(λ-A) = n - P(λI-A)$ – durian Sep 11 '14 at 19:25
• If the matrix has rank $1$, that means the dimension of its kernel is $n-1$, i.e. $0$ is an eigenvalue of geometric multiplicity $n-1$. Since $A$ is symmetric, $0$ also has algebraic multiplicity $n-1$. Thus, the remaining eigenvalue $\lambda = n$ has (algebraic and geometric) multiplicity $1$. – BaronVT Sep 12 '14 at 19:35
• Compare this to the following "non-example": $B = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$. Here, the matrix has rank $1$, so $0$ is still an eigenvalue of geometric multiplicity $1$, but has algebraic multiplicity $2$, so you can't use the same "trick" to write the characteristic polynomial. – BaronVT Sep 12 '14 at 19:40

For A you can use the fact that if the sum of all the rows is equal, then this sum, $n$ is eigenvalue, and since the rank is 1, it's mean that $\dim\ker A = n-1$ which means that $0$ is eigenvalue from order $n-1$.

from that you can conclude that the characteristic polynomial of $A$ is $$p(\lambda)=\lambda ^ {n-1}(\lambda -n)$$