Linear Algebra and Its Applications, Exercise 3.3.10

Exercise 3.3.10. Given mutually orthogonal vectors $a_1$ , $a_2$ , and $b$ and the matrix $A$ with columns $a_1$ and $a_2$ , what are $A^TA$ and $A^Tb$ ? What is the projection of $b$ onto the plane formed by $a_1$ and $a_2$ ?

Answer: We have

$A^TA = \begin{bmatrix} a_1^T \\ a_2^T \end{bmatrix} \begin{bmatrix} a_1&a_2 \end{bmatrix} = \begin{bmatrix} a_1^Ta_1&a_1^Ta_2 \\ a_2^Ta_1&a_2^Ta_2 \end{bmatrix} = \begin{bmatrix} \|a_1\|^2&0 \\ 0&\|a_2\|^2 \end{bmatrix}$

where the zero entries are the result of $a_1$ and $a_2$ being orthogonal.

Similarly we have

$A^Tb = \begin{bmatrix} a_1^T \\ a_2^T \end{bmatrix} b = \begin{bmatrix} a_1^Tb \\ a_2^Tb \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} = 0$

where the zero entries are the result of $a_1$ and $a_2$ being orthogonal to $b$ .

Since $b$ is orthogonal to both $a_1$ and $a_2$ it is orthogonal to any linear combination of $a_1$ and $a_2$ and therefore is orthogonal to the plane spanned by $a_1$ and $a_2$ . The projection of $b$ onto that plane is therefore the zero vector.

This also follows from the formula for the projection matrix $P$ corresponding to the matrix $A$ . The projection $p$ of $b$ onto the column space of $A$ (the space spanned by $a_1$ and $a_2$ ) is

$p = Pb = A(A^TA)^{-1}A^Tb = A(A^TA)^{-1} \cdot 0 = 0$

since $A^Tb = 0$ as discussed above.

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, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

Linear Algebra and Its Applications, Exercise 3.3.10

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