19.9. EXERCISES 425
Example 19.8.7 Here is a symmetric matrix which has eigenvalues 6,−12,18
A =
1 −4 13−4 10 −413 −4 1
Find a matrix U such that UT AU is a diagonal matrix.
From the above explanation the columns of this matrix U are eigenvectors of unit lengthand in fact this is sufficient to obtain the matrix. After doing row operations to find theeigenvectors and then dividing each by its magnitude, you obtain 1 −4 13
−4 10 −413 −4 1
16
√6
13
√6
16
√6
=
√
62√
6√6
= 6
16
√6
13
√6
16
√6
1 −4 13
−4 10 −413 −4 1
− 12
√2
012
√2
=
6√
20
−6√
2
=−12
− 12
√2
012
√2
1 −4 13
−4 10 −413 −4 1
13
√3
− 13
√3
13
√3
=
6√
3−6
√3
6√
3
= 18
13
√3
− 13
√3
13
√3
Thus the matrix of interest is
U =
16
√6 − 1
2
√2 1
3
√3
13
√6 0 − 1
3
√3
16
√6 1
2
√2 1
3
√3
Then 1
6
√6 − 1
2
√2 1
3
√3
13
√6 0 − 1
3
√3
16
√6 1
2
√2 1
3
√3
T 1 −4 13−4 10 −413 −4 1
16
√6 − 1
2
√2 1
3
√3
13
√6 0 − 1
3
√3
16
√6 1
2
√2 1
3
√3
=
6 0 00 −12 00 0 18
19.9 Exercises1. Let
{u1, · · · ,un}
be a basis for Fn and define a mapping T : Fn → span(v1, · · · ,vr) as follows.
T
(n
∑k=1
akuk
)≡
r
∑k=1
akvk
Explain why this is a linear transformation.