Linear Algebra II (pdf, 500 kB)

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the standard basis vectors. However, Lemma 8.18 tells us that (for example)
v1 = (1, 1, 0) ⊤ will do. So we make a first change of basis to obtain
A ′ ⎛
1
= ⎝0 1
1
⎞ ⎛
0 1
0⎠
A ⎝1 0
1
⎞ ⎛
0 2
0⎠
= ⎝1 1
0
⎞
2
1⎠
.
0 0 1 0 0 1 2 1 0
Now we have to find a basis of the orthogonal complement L(v1) ⊥ . This can be
done by adding suitable multiples of v1 to the other basis elements, in order to
make the off-diagonal entries in the first row and column of the matrix zero. Here
we have to add −1/2 times the first basis vector to the second, and add −1 times
the first basis vector to the third. This gives
A ′′ ⎛
1 0 0
= ⎝− 1
⎞
1 0⎠
2 A
−1 0 0
′
⎛
1 1 − 2
⎝ −1
⎞ ⎛
2 0 0
0 1 0 ⎠ = ⎝0 −
0 0 1
1
⎞
0 ⎠
2 .
0 0 −2
We are lucky: this matrix is already diagonal. (Otherwise, we would have to
continue in the same way with the 2 × 2 matrix in the lower right.) The total
change of basis is indicated by the product of the two P ’s that we have used:
⎛ ⎞ ⎛
1
1 0 0 1 − 2
P = ⎝1 1 0⎠
⎝
0 0 1
−1
⎞ ⎛
1 1 − 2
0 1 0 ⎠ = ⎝
0 0 1
−1
1 1 2 −1
⎞
⎠
0 0 1
so the desired basis is v1 = (1, 1, 0) ⊤ , v2 = (− 1 1 , 2 2 , 0)⊤ , v3 = (−1, −1, 1) ⊤ .
For algebraically closed fields like C, we get a very nice result.
8.22. Theorem (Classification of Symmetric Bilinear Forms Over C). Let
F be algebraically closed, for example F = C. Then every symmetric matrix
A ∈ Mat(n, F ) is congruent to a matrix
Ir 0
,
0 0
and the rank 0 ≤ r ≤ n is uniquely determined.
Proof. By Thm. 8.19, A is congruent to a diagonal matrix, and we can assume that
all zero diagonal entries come at the end. Let ajj be a non-zero diagonal entry.
Then we can scale the corresponding basis vector by 1/ √ ajj (which exists in F ,
since F is algebraically closed); in the new matrix we get, this entry is then 1.
The uniqueness statement follows from the fact that n − r is the dimension of the
(left or right) kernel of the associated bilinear form.
If F = R, we have a similar statement. Let us first make a definition.
8.23. Definition. Let V be a real vector space, φ a symmetric bilinear form on V.
Then φ is positive definite if
φ(v, v) > 0 for all v ∈ V \ {0}.
8.24. Remark. A positive definite symmetric bilinear form is non-degenerate: if
v = 0, then φ(v, v) > 0, so = 0, hence v is not in the (left or right) kernel of v.
For example, this implies that the Hilbert matrix from Example 8.13 is invertible.