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Given:

$A \in M_{n \times n} (\mathbb R)$, $n\geq 2$, rank($A) = 1$, trace($A) = 0$. Prove A is not diagonalizable and find $P_A(x)$.

So I said:

if $n \geq 2$ and rank($A)=1$ then $A$ is not invertible. That means that it has an eigenvalue 0. Its' geometric multiplicity is $n-1$, since, again, rank$(A)=1$. Now we also know that the trace is the sum of all the eigenvalues which means $\operatorname{trace}(A) = 0^{n-1} + \lambda_x = \lambda_x = 0$ which means ALL the eigenvalues are $0$.

Is that correct? and if so, does it mean that it is not diagonalizable? (and if its correct then of course $P_A(x) = \lambda^{n}$.

learner
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TheNotMe
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2 Answers2

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Yes, it's correct so far.

But probably it is better to say the roots of characteristic polynomial instead of the eigenvalues, and you should note somewhere that they may also be complex numbers.

There is also one tiny error/mistype: it should be ${\rm trace}(A)=(n-1)\cdot 0+\lambda_x$.

Yes, we have that the characteristic polynomial is $P_A(x)=x^n$.

And, for finishing, either you can consider the Jordan form of $A$: it must have only $0$'s in the diagonal... (A small example is $A=\pmatrix{0&1\\0&0}$.)

Or, you can conclude that such a diagonalizable matrix could be only ${\rm diag}(0,0,0,0,..,0)$ which is the null matrix, but this has rank $0$.

Berci
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  • "Better to say the roots of characteristic polynomial instead of the eigenvalues" - why do you say this? – Cocopuffs Jun 01 '13 at 17:15
  • @Cocopuffs because just saying "eigenvalues" does not take into account multiplicity. – Ragib Zaman Jun 01 '13 at 21:25
  • I'd rather take the last conclusion. But can you please explain how to conclude such a thing that if it was diagonlizable it should be the null matrix? Is it the only one that is diagonalizable and has that characteristic polynomial? – TheNotMe Jun 02 '13 at 08:16
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If $A$ diagonalized to $\rm{diag} (x_1,\cdots, x_n)$ then $\rm{rank} (A)=1$ implies $x_ix_j=0$ for all $i\neq j .$ If any $x_k$ is not $0$ then all the other $x_i$ are, so $\rm trace (A) = x_k\neq 0.$

Ragib Zaman
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  • please explain why $rank(A)=1$ implies $x_ix_j =0$. – TheNotMe Jun 02 '13 at 08:13
  • Read this for many equivalent definitions of the rank. Look at the alternative definitions section, I am using the fact that the rank is the size of the largest non-vanishing minor. Here the rank is 1 so every 2x2 submatrix has determinant zero. $x_i x_j$ is the determinant of the 2x2 submatrix obtained by deleting every row except the i-th and j-th, and deleting every column except the i-th and j-th. – Ragib Zaman Jun 02 '13 at 11:17
  • But, if $A$ is diagonalizable, and its' rank is $1$, does that mean that it preserves the rank even when its' in its' diagonal form? – TheNotMe Jun 02 '13 at 14:00
  • @TheNotMe Yes, the rank of two conjugate matrices is equal. $\text{diag}(x_1,\cdots, x_n) $ is essentially the same abstract linear operator as $A$ but from a different choice of basis, so the rank, being the dimension of the image, is the same. If you haven't seen this viewpoint before, you might be able to directly show that the dimension of the column space is the same directly, but might be best just to familiarize yourself with the definition as the dimension of the image. – Ragib Zaman Jun 02 '13 at 14:57
  • One more thing. They said that $tr(A) = 0$ before we started making it diagonal. And I am pretty sure the same $trace$ does not hold after we diagonalize it. – TheNotMe Jun 02 '13 at 16:17
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    @TheNotMe It's amazing but it actually does ;). The trace is the sum of the eigenvalues, i.e the coefficient of $x^{n-1}$ in $\det(A-\lambda I)=0.$ Letting $D=\text{diag}(x_1, \cdots, x_n)$ so $A= P^{-1} D P$ we have $ \det( A - \lambda I ) = \det( P^{-1} D P - P^{-1} \lambda I P) = \det(P^{-1}) \det(D - \lambda I) \det(P) = \det(D - \lambda I)$ so $A$ has the same characteristic polynomial as $D$, so they have the same eigenvalues, trace, determinant, etc. – Ragib Zaman Jun 02 '13 at 16:19
  • Thank you so much for this helpful post and comments. Much appreciated! – TheNotMe Jun 02 '13 at 16:24
  • @TheNotMe No worries. – Ragib Zaman Jun 02 '13 at 16:27
  • According to your way, how can we find what the characteristic polynomial is? – TheNotMe Jun 02 '13 at 18:02
  • @TheNotMe What do you mean? The definition of the characteristic polynomial of a matrix A (in variable $\lambda$) is $\det(A-\lambda I).$ – Ragib Zaman Jun 03 '13 at 00:30