Linear Algebra and Its Applications, Exercise 3.3.14

Exercise 3.3.14. Find the projection matrix $P$ onto the plane spanned by the vectors $(1, 1, 0)$ and $(1, 1, 1)$ . Find a nonzero vector $b$ that $P$ projects to zero.

Answer: The plane in question is the column space of the matrix

$A = \begin{bmatrix} 1&1 \\ 1&1 \\ 0&1 \end{bmatrix}$

The projection matrix $P = A(A^TA)^{-1}A^T$ . We have

$A^TA = \begin{bmatrix} 1&1&0 \\ 1&1&1 \end{bmatrix} \begin{bmatrix} 1&1 \\ 1&1 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 2&2 \\ 2&3 \end{bmatrix}$

so that its inverse is

$(A^TA)^{-1} = \frac{1}{2 \cdot 3 - 2 \cdot 2} \begin{bmatrix} 3&-2 \\ -2&2 \end{bmatrix}$

$= \frac{1}{2} \begin{bmatrix} 3&-2 \\ -2&2 \end{bmatrix} = \begin{bmatrix} \frac{3}{2}&-1 \\ -1&1 \end{bmatrix}$

We then have

$P = A(A^TA)^{-1}A^T$

$= \begin{bmatrix} 1&1 \\ 1&1 \\ 0&1 \end{bmatrix} \begin{bmatrix} \frac{3}{2}&-1 \\ -1&1 \end{bmatrix} \begin{bmatrix} 1&1&0 \\ 1&1&1 \end{bmatrix}$

$= \begin{bmatrix} 1&1 \\ 1&1 \\ 0&1 \end{bmatrix} \begin{bmatrix} \frac{1}{2}&\frac{1}{2}&-1 \\ 0&0&1 \end{bmatrix}$

$= \begin{bmatrix} \frac{1}{2}&\frac{1}{2}&0 \\ \frac{1}{2}&\frac{1}{2}&0 \\ 0&0&1& \end{bmatrix}$

This completes the first part of the exercise. For the second part, we need to find a nonzero vector $b$ projected into zero by $P$ . This will be true if $b$ is in the left nullspace of $A$ , since $b$ will then be orthogonal to the plane corresponding to the column space of $A$ .

We therefore look for a vector in the left nullspace $\mathcal{N}(A^T)$ by solving the system $A^Ty = 0$ :

$\begin{bmatrix} 1&1&0 \\ 1&1&1 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} = 0$

We perform Gaussian elimination by multiplying the first row by 1 and subtracting it from the second row:

$\begin{bmatrix} 1&1&0 \\ 1&1&1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&1&0 \\ 0&0&1 \end{bmatrix}$

The resulting matrix is in echelon form, with pivots in columns 1 and 3. Thus $y_1$ and $y_3$ are basic variables, and $y_2$ is a free variable. Setting $y_2 = 1$ we have $y_3 = 0$ (from the second row) and $y_1 = -1$ (from the first row). The vector $b = (-1, 1, 0)$ is therefore a solution to $A^Ty = 0$ and is a basis for the left nullspace of $A$ .

Since $b$ is in $\mathcal{N}(A^T)$ it is orthogonal to $\mathcal{R}(A)$ and is therefore projected by $P$ into zero:

$Pb = \begin{bmatrix} \frac{1}{2}&\frac{1}{2}&0 \\ \frac{1}{2}&\frac{1}{2}&0 \\ 0&0&1& \end{bmatrix} \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fifth Edition and the accompanying free online course, and Dr Strang’s other books.

Linear Algebra and Its Applications, Exercise 3.3.14

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Linear Algebra and Its Applications, Exercise 3.3.14

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