Linear Algebra and Its Applications, Review Exercise 2.25

Posted on December 28, 2013 by hecker

Review exercise 2.25. Suppose that T is a linear transformation from \mathbb{R}^3 to itself, and T transforms the point (u, v, w) to the point (u+v+w, u+v, u). What does the inverse transformation T^{-1} do to the point (x, y, z)?

Answer: The effect of T^{-1} is to reverse the effect of T. Since T takes the first entry of a vector and makes it the last entry of the resulting vector, T^{-1} must take the last entry of a vector and make it the first. So applying T^{-1} to the point (x, y, z) results in a point whose first entry is z.

Next, T takes the second entry of a vector, adds to it the first entry, and makes the  sum the second entry of the resulting vector. In reversing this T^{-1} must take the second entry and subtract from it the original first entry. So applying T^{-1} to the point (x, y, z) results in a point whose second entry is y-z (since z was the original first entry, as discussed in the previous paragraph).

Finally, T takes the third entry of a vector, adds to it the first and second entries, and makes the sum the first entry of the resulting vector . In reversing this T^{-1} must take the first entry and subtract from it the original first entry and second entry. So applying T^{-1} to the point (x, y, z) results in a point whose third entry is x - z - (y-z) = x - y. (Recall that z was the original first entry, and y-z the original second entry.)

The inverse transformation T^{-1} thus transforms the point (x, y, z) into the point (z, y-z, x-y). To confirm this, we apply the transformation T to (z, y-z, x-y) resulting in the point

(z +y-z + x-y, z + y-z, z) = (x, y, z)

The transformation T^{-1} is thus indeed the inverse of the transformation T.

Note that another way to compute T^{-1} is to take the matrix [T] corresponding to the transformation T and compute its inverse [T]^{-1}.

The linear transformation T corresponds to the matrix

[T] = \begin{bmatrix} 1&1&1 \\ 1&1&0 \\ 1&0&0 \end{bmatrix}

so that applying T to (u, v, w) gives

\begin{bmatrix} 1&1&1 \\ 1&1&0 \\ 1&0&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} u+v+w \\ u+v \\ w \end{bmatrix}

We can compute the inverse of [T] using Gauss-Jordan elimination. Start with

\begin{bmatrix} 1&1&1&\vline&1&0&0 \\ 1&1&0&\vline&0&1&0 \\ 1&0&0&\vline&0&0&1 \end{bmatrix}

Subtract 1 times the first row from the second row:

\Rightarrow \begin{bmatrix} 1&1&1&\vline&1&0&0 \\ 0&0&-1&\vline&-1&1&0 \\ 1&0&0&\vline&0&0&1 \end{bmatrix}

Subtract 1 times the first row from the third row:

\Rightarrow \begin{bmatrix} 1&1&1&\vline&1&0&0 \\ 0&0&-1&\vline&-1&1&0 \\ 0&-1&-1&\vline&-1&0&1 \end{bmatrix}

Exchange the second and third rows:

\Rightarrow \begin{bmatrix} 1&1&1&\vline&1&0&0 \\ 0&-1&-1&\vline&-1&0&1 \\ 0&0&-1&\vline&-1&1&0 \end{bmatrix}

Subtract 1 times the third row from the second row:

\Rightarrow \begin{bmatrix} 1&1&1&\vline&1&0&0 \\ 0&-1&0&\vline&0&-1&1 \\ 0&0&-1&\vline&-1&1&0 \end{bmatrix}

Subtract -1 times the third row from the first row:

\Rightarrow \begin{bmatrix} 1&1&0&\vline&0&1&0 \\ 0&-1&0&\vline&0&-1&1 \\ 0&0&-1&\vline&-1&1&0 \end{bmatrix}

Subtract -1 times the second row from the first row:

\Rightarrow \begin{bmatrix} 1&0&0&\vline&0&0&1 \\ 0&-1&0&\vline&0&-1&1 \\ 0&0&-1&\vline&-1&1&0 \end{bmatrix}

Multiply both the second row and the third row by -1:

\Rightarrow \begin{bmatrix} 1&0&0&\vline&0&0&1 \\ 0&1&0&\vline&0&1&-1 \\ 0&0&1&\vline&1&-1&0 \end{bmatrix}

We thus have

[T]^{-1} = \begin{bmatrix} 0&0&1 \\ 0&1&-1 \\ 1&-1&0 \end{bmatrix}

so that

\begin{bmatrix} 0&0&1 \\ 0&1&-1 \\ 1&-1&0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} z \\ y-z \\ x-y \end{bmatrix}

and

[T][T]^{-1} = \begin{bmatrix} 1&1&1 \\ 1&1&0 \\ 1&0&0 \end{bmatrix} \begin{bmatrix} 0&0&1 \\ 0&1&-1 \\ 1&-1&0 \end{bmatrix}

= \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} = I

So the linear transformation T^{-1} as defined by [T]^{-1} is indeed the inverse of the original transformation T.

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.

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