-1
$\begingroup$

Consider the eigenvalue equation $AX=\lambda X$ where A is a $n\times n$ square matrix and $\lambda$ is the eigenvalue corresponding to the eigenvector $\lambda$. For two square matrices, A and $\det(AB)=\det A.\det B$. How to take the determinant of both sides of the equation $(A-\lambda I)X=0$ where $A-\lambda I$ is a $n\times n$ matrix and $X$ is a column vector.

My question is how to show that for nontrivial $X$, $\det(A-\lambda I)=0$? This is a matrix equation and not an algebraic equation of the form $ax=0$.

$\endgroup$
  • 1
    $\begingroup$ You call that "an eigenvalue equation": is $\;\lambda\;$ an eigenvalue of matrix $\;A\;$ ? And you can only take the determinant of a square matrix, not of a column vector which is not a $\;1\times 1\;$ one... $\endgroup$ – DonAntonio Feb 12 '18 at 18:50
  • 1
    $\begingroup$ You don't "take the determinant of both sides of an equation". You want to compute the determinant of $A -\lambda I$, which will be a polynomial in $\lambda$, in order to find the roots. Those roots will be the scalars for which there is an $X$ annihilated by $A=\lambda I$. $\endgroup$ – Ethan Bolker Feb 12 '18 at 18:54
  • $\begingroup$ @EthanBolker So the point is $A-\lambda I$ must be a null matrix, and therefore, have determinant zero. Right? $\endgroup$ – mithusengupta123 Feb 12 '18 at 19:10
  • 2
    $\begingroup$ @mithusengupta123 No, of course not. That means $\;A-\lambda I\;$ , in case $\;\lambda\;$ is an eigenvalue of $\;A\;$ , is a singular, or non-regualr, or non-invertible, matrix...waaay a different thing. $\endgroup$ – DonAntonio Feb 12 '18 at 19:12
  • $\begingroup$ Yes, determinant $0$. No, not the null matrix. $\endgroup$ – Ethan Bolker Feb 12 '18 at 19:21
0
$\begingroup$

Observe that for any real or complex $\;\lambda\;$ and some square matrix $\;A\;$ , we have that $\;\lambda\;$ is an eigenvalue of $\;A\;$ iff $\;A-\lambda I\;$ is a regular matrix iff $\;\det (A-\lambda I)=0\;$ .

You can't take determinant of the neither side in $\;(A-\lambda I)X=0\;$ unless both sides sides are $\;1\times 1\;$ vectors. Determinant is defined only for square matrices .

Now, for an unknown $\;x\;$ , we have that $\;\det (A-xI)=p_A(x)\;$ is the characteristic polynomial of $\;A\;$ ,and we know that its roots are exactly the eigenvalues of $\;A\;$ , so for a number $\;\lambda\;$ we get what I wrote in the first parraph above.

$\endgroup$
0
$\begingroup$

Both sides of the equation are column matrices ($n\times 1$ matrices), so they have no determinant defined (except for the trivial case when $n=1$).

Instead, what is argued here is that the fact that $B\cdot X=0$ (with $B=A-\lambda I$) for some $X\neq 0$ (not equal to the null vector). This implies that $B$ is a singular matrix (otherwise $X=0$ would be the only solution) and being singular it has to be $\det(B)=0$.

That's why $\lambda$ is an eigenvector of $A$ $\iff$ $\det(A-\lambda I)=0$.

$\endgroup$
0
$\begingroup$

If I understand the question correctly, you are asking:

Why $(A-\lambda I)X=0$ for a nontrivial $X$ implies $\det(A-\lambda I)=0?$

Here is why:

If there is a nontrivial $X$ such that $(A-\lambda I)X=0$, then X is an eigenvector of the matrix $(A-\lambda I)$ with the eigenvalue zero. If one of the eigenvalues is zero, the product of the eigenvalues is also zero.

Now, we use the fact that the determinant of every matrix is the product of the eigenvalues of that matrix (see here). Applying this statement to the matrix $(A-\lambda I)$, we have, $\det(A-\lambda I)=0.$

Note: The above statement about determinant of a matrix being the product of its eigenvalues is my favorite definition of determinant. There are several definitions of determinant, and it is not trivial to see how they are equivalent. In my opinion, this statement is really about what determinant is as opposed to how it is calculated or what its properties are. I also find it the most helpful in proofs.

$\endgroup$
  • $\begingroup$ That determinant isn't the product of the eigenvalues, it's a polynomial in $\lambda$. $\endgroup$ – Ethan Bolker Feb 12 '18 at 19:25
  • $\begingroup$ @EthanBolker Determinant of any matrix is the product of the eigenvalues of that matrix (see en.wikipedia.org/wiki/…). Here, I'm talking about the eigenvalues of $A-\lambda I$ not the eigenvalues of $A$. $\endgroup$ – stochastic Feb 12 '18 at 19:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.