Unit four
Section 4.1
Recall that
v \perp w \iff |v+w|^2 = |v|^2 + |w|^2
Which only occurs when
v\cdot w = 0
As
|v+w|^2 = (v + w) \cdot (v + w) = v \cdot v + 2(v\cdot w) + w\cdot w
Because the dot product is linear, we know that V^\perp is a subspace of \mathbb{R}^n. For A (m\times n), \text{Null}(A) = \text{RowSp}(A)^\perp.
The dimension of V\perp is equal to the dimension of n - dim V.
We can see this forming
\begin{align*} A = \begin{pmatrix} & \cdots & v_1^T \end{pmatrix} \end{align*}
Suppose that V is a subspace of \mathbb{R}^\ell of dimension d.
If v_1,\ldots,v_d are linearly independent in V, then they are a basis of V. If the vectors span V, they also form a basis.
This is because any LI subset of V can be enlarged into a basis. Thus the collection of d elements in V clearly forms a basis for V. Replace enlarge with shrink and you have a second proof.
If V \cap W = \{0\},then the dimesnions of v + w is the dimesnion of v plus the dimension of w. The union of any two bases is a basis. Thus the set \{v_1,w_e\} Spans V + W
Suppose c_1v_1 \cdots c_d v_d + b_1 w _ e + \cdots b_ew_e.
Remember that matrix theory doesn’t work over complex numbers.
For problem 14 on the homework, assume that v and w are subspaces of \mathbb{R}^\ell. Find the basis of V \ cap W.
Form A such that the column space is V + W, and the rank being the dimensoo. We can construct the Nullspace by using a basis. Thus A is a
We can take a basis of V \+ W amd send it to a basis in order to get a result of dimension V W = the dimension of the Null space of A. The rank of V + w + V W = dim V plus dim W,=.
Section 4.2 Projections
Four subspaces for the rank 1 simple Matrix
A = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}
A three dimensional shadow can be visualized A = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{pmatrix}
The projection of b onto the line created by the vector a is
P_ab = \mu a
For some \mu. Note that because the perpindicular lines are zero.
a \cdot P_ab = a \cdot \mu a
Therefore
P_ab = \frac{a \cdot b}{a \cdot a}a = \frac{a a^T b}{a^Ta}
P_a = \frac{aa^T}{a^Ta}
Let a = (1,2)
P_a = \frac{1}{5}\{\}
For homework problems 5,7, if a_1 \perp a_2, then P_{a_1} + P_{a_2} is equal to the projection of the span. In \mathbb{R}^3, if you have three orthogonal vectors, then the three projections are equal to the identity matrix.
Section 4.5
Suppose we have a matrix A_{m,n}. This acts as a function mapping \mathbb{R}^n to \mathbb{R}^m.
Note that when we multiply A^+A, we get an n\times n matrix. Lets look at what happens to each subspace under A^+A.
The Null space disapears, as A maps \text{NullSp}(A) \to 0. The row space is restored. Thus we have an interesting map
A^+A|_{\text{Row(A)}} = P_{\text{Row}(A)}
We also have
AA^A|_{\text{Col(A)}} = P_{\text{Col}(A)}
Somewhat suprisingly, (A^+)^+ = A. Every subspace of A is the same under A^T and A^+.