I have a quick question regarding the method of finding eigenvalues and eigenvectors with linear transformations.

Suppose I have a linear transformation T that has a matrix representation A with respect to some basis.

What would $I$ be in $$det(A - \lambda I)$$

Would it be simply the standard identity matrix or the basis vectors for the matrix representation of T?

  • 1
    $\begingroup$ Note that if we express $A - \lambda I$ in terms of some new basis, the result is of the form $S(A - \lambda I)S^{-1}$ for some invertible matrix $S$. This simplifies to $SAS^{-1} - \lambda SIS^{-1} = SAS^{-1} - \lambda I$, in other words, the identity map has the same matrix $I$ no matter what basis we choose. $\endgroup$ – Bungo Apr 4 '18 at 5:45
  • $\begingroup$ @Bungo You should make that an answer instead of a comment. $\endgroup$ – amd Apr 4 '18 at 6:21
  • $\begingroup$ @amd I wasn't sure if this addressed the OP's point of confusion, so I thought a comment would be more suitable. $\endgroup$ – Bungo Apr 4 '18 at 6:25

It is just the identity matrix. That is, the matrix with $1$'s along the diagonal and $0$'s elsewhere.

Recall that this comes from starting with an expression of the form $Av = \lambda v$ and then manipulating this to $Av - \lambda v = 0 \implies (A - \lambda I)v = 0$. Where $I$ is the identity matrix. We see that the basis chosen to produce $A$ has no effect whatsoever on the introduction of $I$ into this expression.

  • $\begingroup$ thanks! appreciate the help friend $\endgroup$ – Question asker Apr 4 '18 at 5:26

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