A linear map has matrix A = pmatrix 6&3 3&6 pmatrix in the standard basis

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Linear algebra and vector spaces · Linear algebra and vector spaces

Conditions of Validity, Ch Vector spaces, linear maps and the dual: the Riesz theorem

Problem

A linear map

T:\mathbb{R}^2\to\mathbb{R}^2

has matrix

\mathsf{A}=\begin{pmatrix}6&3\\3&6\end{pmatrix}

in the standard basis

(1,0),(0,1)

. Someone argues: "

\mathsf{A}

is symmetric, so it represents a symmetric operator; the eigenvalues, which are

9

and

3

, are found in every basis, so in every basis the matrix of

T

must again be symmetric, and in fact

T

'really is' the diagonal matrix

\mathrm{diag}(9,3)

, whatever basis is used to write it down." ((a)) Compute the matrix of

T

in the new basis

\mathbf{e}'_1=(1,0)

\mathbf{e}'_2=(2,1)

. ((b)) State which of trace, determinant, eigenvalues and symmetry survive the change of basis, using the two matrices as evidence. ((c)) Find the specific error(s) in the quoted argument.

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