### Orthogonal and symmetric maps: Symmetric maps

### Orthonormal bases and symmetric maps

From the *previously proven properties of symmetric matrices* we now derive their *diagonalizability*.

Diagonalizability of symmetric maps

Let #V# be an inner product space of finite dimension and let #L:V\rightarrow V# be a symmetric linear map. Then there exists an orthonormal basis for #V# consisting of eigenvectors of #L#.

In terms of matrices the result can be formulated as follows.

Diagonalizability of symmetric matrices Let #n# be a natural number and #A# a symmetric #(n\times n)#-matrix. Then there is an orthogonal matrix #X# such that #X\, A\, X^{-1}# is a diagonal matrix.

#\alpha = # #\basis{\left[ {{1}\over{\sqrt{3}}} , -{{1}\over{\sqrt{3}}} , -{{1}\over{\sqrt{3}}} \right] , \left[ {{1}\over{\sqrt{2}}} , 0 , {{1}\over{\sqrt{2}}} \right] , \left[ -{{1}\over{\sqrt{6}}} , -{{\sqrt{2}}\over{\sqrt{3}}} , {{1}\over{\sqrt{6}}} \right] } #

The characteristic polynomial of #A# is \[-x^3+2 x^2+24 x\] This factors as follows into linear polynomials: \[p_A(x) = \left(6-x\right)\cdot x\cdot \left(x+4\right)\] Thus, the symmetric matrix #A# has eigenvalues #0#, #-4#, # 6#. Corresponding eigenvectors are

\[

E_{0}=\linspan{\left[ 1 , -1 , -1 \right] },\ E_{-4}=\linspan{\left[ 1 , 0 , 1 \right] },\ E_{6}=\linspan{\left[ -1 , -2 , 1 \right] }

\] These eigenvectors are perpendicular to each other. Therefore, the choice of a vector of length #1# in each of #E_{0}#, #E_{-4}#, #E_{6}# leads to an orthonormal basis of eigenvectors of #A#:

\[

\alpha = \basis{\left[ {{1}\over{\sqrt{3}}} , -{{1}\over{\sqrt{3}}} , -{{1}\over{\sqrt{3}}} \right] ,\ \left[ {{1}\over{\sqrt{2}}} , 0 , {{1}\over{\sqrt{2}}} \right] ,\ \left[ -{{1}\over{\sqrt{6}}} , -{{\sqrt{2}}\over{\sqrt{3}}} , {{1}\over{\sqrt{6}}} \right] }

\]

The characteristic polynomial of #A# is \[-x^3+2 x^2+24 x\] This factors as follows into linear polynomials: \[p_A(x) = \left(6-x\right)\cdot x\cdot \left(x+4\right)\] Thus, the symmetric matrix #A# has eigenvalues #0#, #-4#, # 6#. Corresponding eigenvectors are

\[

E_{0}=\linspan{\left[ 1 , -1 , -1 \right] },\ E_{-4}=\linspan{\left[ 1 , 0 , 1 \right] },\ E_{6}=\linspan{\left[ -1 , -2 , 1 \right] }

\] These eigenvectors are perpendicular to each other. Therefore, the choice of a vector of length #1# in each of #E_{0}#, #E_{-4}#, #E_{6}# leads to an orthonormal basis of eigenvectors of #A#:

\[

\alpha = \basis{\left[ {{1}\over{\sqrt{3}}} , -{{1}\over{\sqrt{3}}} , -{{1}\over{\sqrt{3}}} \right] ,\ \left[ {{1}\over{\sqrt{2}}} , 0 , {{1}\over{\sqrt{2}}} \right] ,\ \left[ -{{1}\over{\sqrt{6}}} , -{{\sqrt{2}}\over{\sqrt{3}}} , {{1}\over{\sqrt{6}}} \right] }

\]

We can write down the matrix of #(L_A)_\alpha# without calculation, because along the diagonal the eigenvalues will appear:

\[

(L_A)_\alpha =\matrix{0 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 6 \\ }

\] By way of verification of the answer, we compose the

\[\begin{array}{rcl} X\, A\, X^{-1} &=&X\, A\, X^{\top} \\ &=& \matrix{{{1}\over{\sqrt{3}}} & -{{1}\over{\sqrt{3}}} & -{{1}\over{\sqrt{3}}} \\ {{1}\over{\sqrt{2}}} & 0 & {{1}\over{\sqrt{2}}} \\ -{{1}\over{\sqrt{6}}} & -{{\sqrt{2}}\over{\sqrt{3}}} & {{1}\over{\sqrt{6}}} \\ }\, \matrix{-1 & 2 & -3 \\ 2 & 4 & -2 \\ -3 & -2 & -1 \\ }\, \matrix{{{1}\over{\sqrt{3}}} & -{{1}\over{\sqrt{3}}} & -{{1}\over{\sqrt{3}}} \\ {{1}\over{\sqrt{2}}} & 0 & {{1}\over{\sqrt{2}}} \\ -{{1}\over{\sqrt{6}}} & -{{\sqrt{2}}\over{\sqrt{3}}} & {{1}\over{\sqrt{6}}} \\ }^{\top} \\ & =& \matrix{0 & 0 & 0 \\ -2^{{{3}\over{2}}} & 0 & -2^{{{3}\over{2}}} \\ -\sqrt{6} & -2\cdot \sqrt{6} & \sqrt{6} \\ } \, \matrix{{{1}\over{\sqrt{3}}} & {{1}\over{\sqrt{2}}} & -{{1}\over{\sqrt{6}}} \\ -{{1}\over{\sqrt{3}}} & 0 & -{{\sqrt{2}}\over{\sqrt{3}}} \\ -{{1}\over{\sqrt{3}}} & {{1}\over{\sqrt{2}}} & {{1}\over{\sqrt{6}}} \\ }\\ &=&\matrix{0 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 6 \\ }\end{array}\]

\[

(L_A)_\alpha =\matrix{0 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 6 \\ }

\] By way of verification of the answer, we compose the

*transition matrix*#X# from #\alpha# to the standard basis and we check whether conjugation by #X# produces the diagonal matrix with #0#, #-4#, #6# on the diagonal:\[\begin{array}{rcl} X\, A\, X^{-1} &=&X\, A\, X^{\top} \\ &=& \matrix{{{1}\over{\sqrt{3}}} & -{{1}\over{\sqrt{3}}} & -{{1}\over{\sqrt{3}}} \\ {{1}\over{\sqrt{2}}} & 0 & {{1}\over{\sqrt{2}}} \\ -{{1}\over{\sqrt{6}}} & -{{\sqrt{2}}\over{\sqrt{3}}} & {{1}\over{\sqrt{6}}} \\ }\, \matrix{-1 & 2 & -3 \\ 2 & 4 & -2 \\ -3 & -2 & -1 \\ }\, \matrix{{{1}\over{\sqrt{3}}} & -{{1}\over{\sqrt{3}}} & -{{1}\over{\sqrt{3}}} \\ {{1}\over{\sqrt{2}}} & 0 & {{1}\over{\sqrt{2}}} \\ -{{1}\over{\sqrt{6}}} & -{{\sqrt{2}}\over{\sqrt{3}}} & {{1}\over{\sqrt{6}}} \\ }^{\top} \\ & =& \matrix{0 & 0 & 0 \\ -2^{{{3}\over{2}}} & 0 & -2^{{{3}\over{2}}} \\ -\sqrt{6} & -2\cdot \sqrt{6} & \sqrt{6} \\ } \, \matrix{{{1}\over{\sqrt{3}}} & {{1}\over{\sqrt{2}}} & -{{1}\over{\sqrt{6}}} \\ -{{1}\over{\sqrt{3}}} & 0 & -{{\sqrt{2}}\over{\sqrt{3}}} \\ -{{1}\over{\sqrt{3}}} & {{1}\over{\sqrt{2}}} & {{1}\over{\sqrt{6}}} \\ }\\ &=&\matrix{0 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 6 \\ }\end{array}\]

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