Math 415 - Applied Linear Algebra
Diagonalization of symmetric matrices
Theorem: A real matrix Ais symmetric if and only if Acan be diagonalized by an orthogonal
matrix, i.e. A=UDU −1with Uorthogonal and Ddiagonal.
To illustrate the theorem, let us diagonalize the following matrix by an orthogonal matrix:
A=
1−1 1
−1 1 −1
1−1 1
.
Here is a shortcut to find the eigenvalues. Note that rows 2 and 3 are multiples of row 1,
which means Ahas nullity 2, so that 0 is an eigenvalue with (algebraic) multiplicity at least 2.
Moreover the sum of the three eigenvalues is tr(A) = 3, so the third eigenvalue must be 3.
Let us find the eigenvectors:
λ1=λ2=0:A−0I=
1−1 1
−1 1 −1
1−1 1
∼
1−1 1
000
000
.
Take v1=
1
1
0
and v2=
0
1
1
. They form a basis of the 0-eigenspace, albeit not an orthonormal
basis. Let us apply Gram-Schmidt to obtain an orthonormal basis. (We call the intermediate
orthogonal vectors wi.)
w1=v1=
1
1
0
u1=w1
kw1k=1
√2
1
1
0
w2=v2−proju1(v2) = v2− hu1, v2iu1=
0
1
1
−1
√2(1) 1
√2
1
1
0
=
−1
2
1
2
1
u2=w2
kw2k=1
√6
−1
1
2
.
λ3= 3 : A−3I=
−2−1 1
−1−2−1
1−1−2
∼
1−1−2
−1−2−1
−2−1 1
∼
1−1−2
0−3−3
0−3−3
∼
1−1−2
0 1 1
0 0 0
∼
1 0 −1
0 1 1
0 0 0
.
1
