# Different eigenvalues of the same linear transformation according different bases

I have a question about linear transformation and eigenvalues.

My question:

Given a linear transformation $T:R^3 \to R^3$, And: $E$ is the standard basis of $R^3$, $B$ is another basis of $R^3$.

Let's denote: $A=[T]^B_B$ .

Let's assume that after gaussian elimination process on $A$ we get a matrix $M$ with one row of $0$'s, and now we calculate the eigenvalues of $M$.

• Are the eigenvalues of $M$ also the eigenvalues of the transformation $T?$

I think yes, because the eigenvalues don't change when you change basis, but the correct answer is no, can someone explain to me why?

By the way, Is it correct to say that we always must to work only with the standard basis of $R^3$ to find the eigenvalues of $T?$ (i.e the eigenvalues of $T$ are the roots of the caracteristic polynomial $P_A = Det(A- \lambda \cdot I)$ where $A=[T]^E_E$ , and $E$ is the standard basis)?

Thanks for help!

• Row operations will change the eigenvalues. Otherwise, every invertible matrix would only have the eigenvalue $1$, since they all can be row-reduced to the identity matrix. – Theo Bendit Jul 26 '18 at 11:39
• In gaussian elimination you change both bases to possibly different ones (e.g. $M = [T]^C_D$, where $C \neq D$). – Stefan Jul 26 '18 at 11:40

For your first question, I want to cite @Theo Bendit's comment and generally add: Gaussian elimination changes many properties of matrices if you're not careful with it, including eigenvalues and e.g. the determinant(at least with in the general way).

For your second question: No, the eigenvalue of the operator $T$ does not depend on the basis chosen, if you calculate the roots of $p_A$ for $A$ being a corresponding representation:

Let $A$ and $B$ be similar, that is they represent the same endomorphism w.r.t. different bases, i.e. $B=CAC^{-1}$ for some invertible $C$(which you might call the change-of-basis matrix). Then

$$B-xI=CAC^{-1}-xI=CAC^{-1}-xCIC^{-1}=C(A-xI)C^{-1}$$

Thus, as the determinant distributes over matrix multiplication, we have

$$p_B=\mathrm{det}(B-xI)=\mathrm{det}(C(A-xI)C^{-1})=\mathrm{det}(C)\cdot\mathrm{det}(A-xI)\cdot\mathrm{det}(C^{-1})=\mathrm{det}(C)\cdot\mathrm{det}(A-xI)\cdot\mathrm{det}(C)^{-1}=\mathrm{det}(A-xI)=p_A$$

The last steps follow from the elementary property of determinants for invertible matrices that $\mathrm{det}(C^{-1})=\mathrm{det}(C)^{-1}$.

EDIT: Note that is makes thus sense to define the characteristic polynomial for an endomorphism, i.e. defining $p_T$ as it made sense to define the determinant for endomorphisms instead of only matrices.