## Announcements:

Today we finish 5.2 and start 5.3. Read Sections 5.3 and 5.4 for Monday. Work through recommended homework questions.

Tutorials: Next week: review.
Office hour: Monday, 1:30-2:30, MC103B.
Help Centers: Monday-Friday 2:30-6:30 in MC 106.

Final exam: Covers whole course, with an emphasis on the material in Chapters 4 and 5 (after the midterm). Our course will end with Section 5.4.

Question: If $W = \R^n$, then $W^\perp = \query{\{ \vec 0 \}}$

T/F: An orthogonal basis $\{ \vv_1, \ldots, \vv_k \}$ must have $$\vv_i \cdot \vv_j = \begin{cases} 0 & \text{if } i \neq j \\ 1 & \text{if } i = j \end{cases}$$

### Review of Section 5.2: Orthogonal Complements and Orthogonal Projections

We saw in Section 5.1 that orthogonal and orthonormal bases are particularly easy to work with. In Section 5.3, we will learn how to find these kinds of bases. In this section, we learn the tools which will be needed in Section 5.3. We will also find a new way to understand the subspaces associated to a matrix.

### Orthogonal Complements

Definition: Let $W$ be a subspace of $\R^n$. A vector $\vv$ is orthogonal to $W$ if $\vv$ is orthogonal to every vector in $W$. The orthogonal complement of $W$ is the set of all vectors orthogonal to $W$ and is denoted $W^\perp$. So $$\kern-4ex W^\perp = \{ \vv \in \R^n : \vv \cdot \vw = 0 \text{ for all } \vw \text{ in } W \}$$

An example to keep in mind is where $W$ is a plane through the origin in $\R^3$ and $W^\perp$ is $\span(\vn)$, where $\vn$ is the normal vector to $W$.

Theorem 5.9: Let $W$ be a subspace of $\R^n$. Then:
a. $W^\perp$ is a subspace of $\R^n$.
b. $(W^\perp)^\perp = W$
c. $W \cap W^\perp = \{ \vec 0 \}$
d. If $W = \span(\vw_1, \ldots, \vw_k)$, then $\vv$ is in $W^\perp$ if and only if $\vv \cdot \vw_i = 0$ for all $i$.

We proved all of these except part (b), which will come today.

Theorem 5.10: Let $A$ be an $m \times n$ matrix. Then $$\kern-4ex (\row(A))^\perp = \null(A) \qtext{and} (\col(A))^\perp = \null(A^T)$$

The first two are in $\R^n$ and the last two are in $\R^m$. These are the four fundamental subspaces of $A$.

### Orthogonal projection

Let $\vu$ be a nonzero vector in $\R^n$, and for any $\vv$ in $\R^n$ define: $$\proj_{\vu}(\vv) = \left( \frac{\vu \cdot \vv}{\vu \cdot \vu} \right) \vu .$$ $$\Perp_{\vu}(\vv) = \vv - \proj_{\vu}(\vv)$$

If we write $W = \span(\vu)$, then $\vw = \proj_{\vu}(\vv)$ is in $W$, $\vw^\perp = \Perp_{\vu}(\vv)$ is in $W^\perp$, and $\vv = \vw + \vw^\perp$. We can do this more generally:

Definition: Let $W$ be a subspace of $\R^n$ and let $\{ \vu_1, \ldots, \vu_k \}$ be an orthogonal basis for $W$. For $\vv$ in $\R^n$, the orthogonal projection of $\vv$ onto $W$ is the vector $$\proj_W(\vv) = \proj_{\vu_1}(\vv) + \cdots + \proj_{\vu_k}(\vv)$$ The component of $\vv$ orthogonal to $W$ is the vector $$\Perp_W(\vv) = \vv - \proj_W(\vv)$$

We will show soon that $\Perp_W(\vv)$ is in $W^\perp$.

Note that multiplying $\vu$ by a scalar in the earlier example doesn't change $W$, $\vw$ or $\vw^\perp$. We'll see later that the general definition also doesn't depend on the choice of orthogonal basis.

### New material

Example: Let $W = \span(\vu_1, \vu_2)$, where $\vu_1 = \collll 1 1 0 0$ and $\vu_2 = \collll 0 0 1 1$. Compute $\proj_W(\vv)$ and $\Perp_W(\vv)$, where $\vv = \collll 1 3 2 4$. On whiteboard.

In the previous example, $\Perp_W(\vv)$ is in $W^\perp$. This is always the case. On whiteboard.

Now we will see that $\proj$ and $\Perp$ don't depends on the choice of orthogonal basis. Here and in the rest of the section, we assume that every subspace has at least one orthogonal basis.

Theorem 5.11: Let $W$ be a subspace of $\R^n$ and let $\vv$ be a vector in $\R^n$. Then there are unique vectors $\vw$ in $W$ and $\vw^\perp$ in $W^\perp$ such that $\vv = \vw + \vw^\perp$.

Proof: We saw above that such a decomposition exists, by taking $\vw = \proj_W(\vv)$ and $\vw^\perp = \Perp_W(\vv)$, using an orthogonal basis for $W$.

We now show that this decomposition is unique. So suppose $\vw = \vw_1 + \vw_1^\perp$ is another such decomposition. Then $\vw + \vw^\perp = \vw_1 + \vw_1^\perp$, so $$\vw - \vw_1 = \vw_1^\perp - \vw^\perp$$ The left hand side is in $W$ and the right hand side is in $W^\perp$ (why?), so both sides must be zero (why?). So $\vw = \vw_1$ and $\vw^\perp = \vw_1^\perp$.$\quad\Box$

Note that $\perp$ is an operation on subspaces, but is not an operation on vectors.

Now we can prove part (b) of Theorem 5.9.

Corollary 5.12: If $W$ is a subspace of $\R^n$, then $(W^\perp)^\perp = W$.

Proof: If $\vw$ is in $W$ and $\vx$ is in $W^\perp$, then $\vw \cdot \vx = 0$. This means that $\vw$ is in $(W^\perp)^\perp$. So $W \subseteq (W^\perp)^\perp$.

We need to show that every vector in $(W^\perp)^\perp$ is in $W$. So let $\vv$ be a vector in $(W^\perp)^\perp$. By the previous result, we can write $\vv$ as $\vw + \vw^\perp$, where $\vw$ is in $W$ and $\vw^\perp$ is in $W^\perp$. Then \kern-4ex \begin{aligned} 0 &= \vv \cdot \vw^\perp = (\vw + \vw^\perp)\cdot\vw^\perp \\ &= \vw \cdot \vw^\perp + \vw^\perp \cdot \vw^\perp = 0 + \vw^\perp \cdot \vw^\perp = \vw^\perp \cdot \vw^\perp \end{aligned} So $\vw^\perp = \vec 0$ and $\vv = \vw$ is in $W$.$\quad\Box$

This next result is related to the Rank Theorem:

Theorem 5.13: If $W$ is a subspace of $\R^n$, then $$\dim W + \dim W^\perp = n$$

Proof: Let $\{ \vu_1, \ldots, \vu_k \}$ be an orthogonal basis of $W$ and let $\{ \vv_1, \ldots, \vv_\ell \}$ be an orthogonal basis of $W^\perp$. Then $\{ \vu_1, \ldots, \vu_k, \vv_1, \ldots, \vv_\ell \}$ is an orthogonal basis for $\R^n$. (Explain.) The result follows.$\quad\Box$

Example: For $W$ a plane in $\R^3$, $2 + 1 = 3$.

The Rank Theorem follows if we take $W = \row(A)$, since then $W^\perp = \null(A)$:

Corollary 5.14 (The Rank Theorem, again): If $A$ is an $m \times n$ matrix, then $$\rank(A) + \nullity(A) = n$$

Note: The logic here can be reversed. We can use the rank theorem to prove Theorem 5.13, and Theorem 5.13 can be used to prove Corollary 5.12.

### The Gram-Schmidt Process

This is a fancy name for a way of converting a basis into an orthogonal or orthonormal basis. And it's pretty clear how to do it, given what we know.

Example: Let $W = \span(\vx_1, \vx_2)$ where $\vx_1 = \colll 1 1 0$ and $\vx_2 = \colll {-2} 0 1$. Find an orthogonal basis for $W$.

Solution: Ideas? Do on whiteboard.

Question: What if we had a third basis vector $\vx_3$?

Theorem 5.15 (The Gram-Schmidt Process): Let $\{ \vx_1, \ldots, \vx_k \}$ be a basis for a subspace $W$ of $\R^n$. Write $W_1 = \span(\vx_1)$, $W_2 = \span(\vx_1, \vx_2)$, $\ldots$, $W_k = \span(\vx_1, \ldots, \vx_k)$. Define: \kern-9ex \begin{aligned} \vv_1 &= \vx_1 \\ \vv_2 &= \Perp_{W_1}(\vx_2) = \vx_2 - \frac{\vv_1 \cdot \vx_2}{\vv_1 \cdot \vv_1} \vv_1 \\ \vv_3 &= \Perp_{W_2}(\vx_3) = \vx_3 - \frac{\vv_1 \cdot \vx_3}{\vv_1 \cdot \vv_1} \vv_1 - \frac{\vv_2 \cdot \vx_3}{\vv_2 \cdot \vv_2} \vv_2 \\ &\vdots \\ \vv_k &= \Perp_{W_{k-1}}(\vx_k) = \vx_k - \frac{\vv_1 \cdot \vx_k}{\vv_1 \cdot \vv_1} \vv_1 - \cdots - \frac{\vv_{k-1} \cdot \vx_k}{\vv_{k-1} \cdot \vv_{k-1}} \vv_{k-1} \end{aligned} Then for each $i$, $\{ \vv_1, \ldots, \vv_i \}$ is an orthogonal basis for $W_i$. In particular, $\{ \vv_1, \ldots, \vv_k \}$ is an orthogonal basis for $W = W_k$.

Explain verbally.