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I can't see how to do this. I cant see how those two vectors can span a subspace of $\mathbb{R}^3$ when they both have 4 components.

Here's what I've done. I tried to find the column space of these two vectors by putting them into a $4\times2$ matrix

$ \left[\begin{array}{rr|r} -4 & 2 & a \\ 2 & 2 & b \\ 2 & 2 & c \\ -4 & 0 & d \end{array}\right] $

I then row-reduced and got the following results.

$ \left[\begin{array}{rr|l} -4 & 2 & a \\ 0 & 6 & 2b - a\\ 0 & 0 & c - b \\ 0 & 0 & 3d - 2a + 2b \end{array}\right] $

So then I used the equation

$3d - 2a + 2b = 0$ to obtain two $1\times3$ vectors that span $\mathbb{R}^3$

$a = b + \frac{3}{2}d$

$\begin{bmatrix} a \\ b \\ d \end{bmatrix} = \operatorname{span} \left\{\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} \frac{3}{2} \\ 0 \\ 1 \end{bmatrix}\right\}$

But I cant project a $4\times1$ onto these vectors because I cant dot a $1\times4$ vector with a $1\times3$ vector.

I have a feeling I have got the workings of this question all wrong. What am I supposed to be doing?

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    Could somebody explain to me what is actually being asked in this question? I think I don't understand what is being asked...2012-04-11

1 Answers 1

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Here's another approach. Let $v_1, v_2$ be the given basis for the subspace $V$. Let $p$ be the desired projection. Since $p \in V$, it can be written as $p=[v_1 v_2] x$, where $x \in \mathbb R^2$. The problem then reduces to finding $x$.

Since $p$ is the projection of $v$ on $V$, then $p-v$ must be perpendicular to $V$, or in other words: $=0$, for $i=1,2$. Substituting for $p$ and rewriting slightly gives: $ v_1^T[v_1 v_2] x = v_1^T v$, $ v_2^T[v_1 v_2] x = v_2^T v$.

This reduces to two simple equations $40 x_1 = -20$, $12 x_2 = 6$, and so we have $x_1=-\frac{1}{2}, x_2 = \frac{1}{2}$. Using $p=[v_1 v_2] x$, we get $p^T = (3,0,0,2)$.