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Is the trace and determinant equal to the sum and product, respectively, of the eigenvalues even if a matrix is not diagonalizable?

The proof I've seen for the trace equaling the sum of the eigenvalues and the determinant equaling the eigenvalues assumes a matrix can be written as $PDP^{-1}$. But if it is not diagonalizable, then we can't possibly write $PDP^{-1}$. So what happens then?

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  • $\begingroup$ I guess you are referring to an $n\times n$ matrix? You can find its jordan normal form and then use properties for determinant and trace that hold for block diagonal matrices. en.wikipedia.org/wiki/Block_matrix $\endgroup$ – Test123 Dec 28 '13 at 16:16
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Yes. If the matrix is not diagonalizable then still it can be written in Jordan canonical form. And trace relation and determinant relation can be determined using this.

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  • $\begingroup$ Thanks. I haven't been familiarised with Jordan canonical form as of yet, but soon. $\endgroup$ – thelionkingrafiki Dec 28 '13 at 16:27
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    $\begingroup$ Jordan canonical form is a great theoretical tool to prove wide range of results. Here is a good link to learn about them [Linear Algebra Notes] (people.math.gatech.edu/~bonetto/teaching/6112-fall13/…) $\endgroup$ – piyush_sao Dec 28 '13 at 23:19
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This answer assumes the matrices are taken over $\mathbb C$.

Yes, the statement is still true even if the matrix isn't diagonalizable.

For the proof you saw it is sufficient that $D$ can be taken to be an upper triangular matrix (and it can be taken in such a way, this is Schur's Decomposition Theorem). This is enough because its diagonal entries will be the eigenvalues of the starting matrix.

Jordan Canonical Form is also sufficient, but Schur's Decomposition is a weaker condition.


For completeness I'll add the proofs here.

Let $n\in \mathbb N$ and $A\in \mathcal M_n(\mathbb C)$. Let $\lambda _1, \ldots ,\lambda _n$ be the eigenvalues of $A$. The characteristic polynomial $p_A(z)$ of $A$ is $\color{grey}{p_A(z)=}(z-\lambda _1)\ldots (z-\lambda _n)$.

Schur's Decomposition guarantees the existence of an invertible matrix $P$ and an upper triangular matrix $U$ such that $A=PUP^{-1}$ and $U$'s diagonal entries are exactly $\lambda _1, \ldots ,\lambda _n$.

Since similarity preserves the characteristic polynomial, it follows that the characteristic polynomial $p_U(z)$ of $U$ is $\color{grey}{p_U(z)=}(z-\lambda _1)\ldots (z-\lambda _n)$, therefore $U$ and $A$ have the same eigenvalues with the same algebraic multiplicity.

From the fact that $U$'s diagonal entries are $\lambda _1, \ldots ,\lambda _n$ it follows that the trace of $U$ is the sum of the eigenvalues of $A$ and the determinant of $U$ is the product of the eigenvalues of $A$.

Trace properties yield the following $$\text{tr}(A)=\text{tr}\left(PUP^{-1}\right)=\text{tr}\left(UP^{-1}P\right)=\text{tr}(U),$$ thus proving that the sum of the eigenvalues of $A$ equals $\text{tr}(A)$.

Similarly for the determinant it holds that $$\det(A)=\det\left(PUP^{-1}\right)=\det\left(P\right)\det\left(U\right)\det\left(P^{-1}\right)=\det(U),$$ hence the product of teh eigenvalues of $A$ equals the determinant of $A$.

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  • $\begingroup$ Thank you for shedding light on the topic. $\endgroup$ – thelionkingrafiki Dec 28 '13 at 16:28

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