I have a question about the following exercise:

Let $f:\mathbb{R}^3\to\mathbb{R}^3$ be the linear transformation such that $$f(x,y,z)=(x-y+2z,-2x+y,x+z).$$ Represent the transformation with respect to the basis $\{(1,1,0),(0,1,-1),(1,1,1)\}$.

What I have done:

I found the images of the basis vectors given:

$$ \begin{align}f(1,1,0) &= (0,-1,1)\\ \\ f(0,1,-1) &= (-3,1,-1)\\ \\ f(1,1,1) &= (2,-1,2) \end{align} $$

and then I found how these vectors can be written as a linear combination of the basis vectors given:

$$\begin{align}(0,-1,1)&=\fbox{0}\cdot (1,1,0)+\fbox{(-1)}\cdot (0,1,-1)+\fbox{0}\cdot (1,1,1) \\ \\ (-3,1,-1)&=\fbox{-6}\cdot (1,1,0)+\fbox{4}\cdot (0,1,-1)\;\;\,\,+\fbox{3}\cdot (1,1,1) \\ \\ (2,-1,2)&=\fbox{3}\cdot (1,1,0)+\fbox{(-3)}\cdot (0,1,-1)+\fbox{-1}\cdot (1,1,1)\end{align}$$

thus the matrix which represents $f$ with respect to this basis should be $$\begin{bmatrix}0 & -6 & 3\\ -1 & 4 & -3\\ 0 & 3 & - 1\end{bmatrix}.$$

Is this correct? Or should I write the matrix with respect to the canonical basis of $\mathbb{R}^3$ and then make the change of basis $C^{-1}AC$?

  • 1
    $\begingroup$ How did you calculate the coefficients in the boxes? Some of them, as currently written, are incorrect. $\endgroup$
    – shoteyes
    Jul 10 at 12:33
  • 2
    $\begingroup$ Both methods are correct, and should give you the same answer – if you do them right. $\endgroup$ Jul 10 at 13:05
  • $\begingroup$ @shoteyes they should be correct now, thank you for pointing that out. For each vector $v$, I set up a system to find out which linear combination of basis vectors is equal to it, ie I found out which $c_1,c_2,c_3$ are such that $c_1 v_1+c_2v_2+c_3v_3=v$ and solved it using Gaussian elimination. $\endgroup$
    – lorenzo
    Jul 10 at 15:48
  • 1
    $\begingroup$ Any thoughts on the comments, or on Dan's answer? $\endgroup$ Jul 12 at 13:47
  • 1
    $\begingroup$ @Gerry Myerson they have all been very useful, I have upvoted the comments and accepted Dan's answer, thank you very much $\endgroup$
    – lorenzo
    Jul 12 at 13:57

This is the same. Maybe the square diagram in the sequel shows in the "simplest" way why.

First i have to say something about the used convention for vectors. Because this is the "canonical impediment" when dealing with base change.

We work with column vectors, and matrices act on them by left multiplication. The linear map of left multiplication with a matrix $A$ will be denoted below (abusively) also by $ A$. So $x$ goes via $A$ to $ A\cdot x=Ax$, displayed as $$ x\overset{A}\longrightarrow Ax\ . $$ "Most of the world" uses column vectors. (Some authors write notes or books (e.g. in Word), and find it handy to use row vectors, so they can be simpler displayed in the book rows. In this case linear maps induced by matrices use the multiplication from the right with such matrices. As long as we need in computations only linear combinations the convention is not so important, but it becomes when we use linear maps induced by matrices.)

We will work in the "category" of (finite dimensional) vector spaces (over $\Bbb R$) with a fixed bases. The space $V:=\Bbb R^3$ comes with the canonical base $\mathcal E=(e_1,e_2,e_3)$, where $e_1,e_2,e_3$ are the columns of the matrix $E$ below, $$ E= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\ . $$ We write this object as $(V,\mathcal E)$. By abuse, we may want to write $(V,E)$ instead.

We start with two objects in this category. For our purposes let them have the same underlying vector space $V=W=\Bbb R^3$, first object is $(V,\mathcal B=(b_1,b_2,b_3))$, and the second object is $(W,\mathcal C=(c_1,c_2,c_3))$.

A linear map $g:V\to W$ is defined "abstractly", and has no need for chosen bases. But in practice, $g$ is usually given bases-specific in the following way. Let $v$ be a vector in $V$. We write it w.r.t. $\mathcal B$ as $v=x_1b_1+x_2b_2+x_3b_3$, and write this data as a column vector: $$ v = \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}_{\mathcal B} :=x_1b_1+x_2b_2+x_3b_3 \ . $$ Then we consider a matrix $M=M_{\mathcal B, \mathcal C}$ and build the matrix multiplication vector: $$ \begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix} = M \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\ . $$ Then we consider the vector $w\in W$ which written in base $\mathcal C$ has the $y$-components, so $$ w = \begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix}_{\mathcal C} :=y_1c_1+y_2c_2+y_3c_3 \ , $$ and the map $g$ is mapping linearly $v$ to $w$.

This concludes the section related to conventions and notations.

Let $\mathcal C$ be the base from the OP, the base with vectors which are columns of $$ C= \begin{bmatrix} 1 & 0 & 1\\ 1 & 1 & 1\\ 0 & -1 & 1 \end{bmatrix}\ . $$ Let $A$ be the matrix for the given linear map $f$ w.r.t. the canonical base $\mathcal E$. $$ A=\begin{bmatrix}1&-1&2\\-2&1&0\\1&0&1\end{bmatrix}\ . $$ Consider now the diagram: $\require{AMScd}$ \begin{CD} (V,E) @>A>f> (V,E) \\ @A C A {\operatorname{id}}A @A {\operatorname{id}}A C A\\ (V,C) @>f>{C^{-1}AC}> (V,C) \\ \end{CD}

Indeed, $C$ is the matrix of the identity seen as a map $(V,\mathcal C)\to(V,\mathcal E)$. For instance, $$ c_1=\begin{bmatrix}1\\0\\0\end{bmatrix}_{\mathcal C} \qquad\text{ goes to }\qquad c_1 =\begin{bmatrix}1\\1\\0\end{bmatrix}_{\mathcal E} =C\begin{bmatrix}1\\0\\0\end{bmatrix}_{\mathcal E}\ . $$

It remains to compute explicitly the matrix $C^{-1}AC$. Computer, of course:

sage: A = matrix(3, 3, [1, -1, 2,  -2, 1, 0,  1,  0, 1])
sage: C = matrix(3, 3, [1,  0, 1,   1, 1, 1,  0, -1, 1])
sage: A
[ 1 -1  2]
[-2  1  0]
[ 1  0  1]
sage: C
[ 1  0  1]
[ 1  1  1]
[ 0 -1  1]
sage: C.inverse() * A  * C
[ 0 -6  3]
[-1  4 -3]
[ 0  3 -1]

And we check the result both ways:


$\require{AMScd}$ \begin{CD} c_j=\begin{bmatrix}1\\1\\0\end{bmatrix}_{\mathcal E}\ ,\ \begin{bmatrix}0\\1\\-1\end{bmatrix}_{\mathcal E}\ ,\ \begin{bmatrix}1\\1\\1\end{bmatrix}_{\mathcal E} @>A>f> fc_j=\begin{bmatrix}0\\-1\\1\end{bmatrix}_{\mathcal E}\ ,\ \begin{bmatrix}-3\\1\\-1\end{bmatrix}_{\mathcal E}\ ,\ \begin{bmatrix}2\\-1\\2\end{bmatrix}_{\mathcal E} \\ @A C A {\operatorname{id}}A @A {\operatorname{id}}A C A\\ c_j=\begin{bmatrix}1\\0\\0\end{bmatrix}_{\mathcal C}\ ,\ \begin{bmatrix}0\\1\\0\end{bmatrix}_{\mathcal C}\ ,\ \begin{bmatrix}0\\0\\1\end{bmatrix}_{\mathcal C} @>f>{C^{-1}AC}> fc_j=\begin{bmatrix}0\\-1\\0\end{bmatrix}_{\mathcal C}\ ,\ \begin{bmatrix}-6\\4\\3\end{bmatrix}_{\mathcal C}\ ,\ \begin{bmatrix}3\\-3\\-1\end{bmatrix}_{\mathcal C} \end{CD}

Or simpler, using block matrices, and ignoring the knowledge of the bases:

$\require{AMScd}$ \begin{CD} C @>A>f> AC \\ @A C A {\operatorname{id}}A @A {\operatorname{id}}A C A\\ E @>f>{C^{-1}AC}> C^{-1}AC \end{CD}

$\bf(2)$ In the spirit of the OP, using copy+pasted+corrected row vector computations:

$$ \begin{aligned} (0,-1,1) &=\boxed{0}\cdot (1,1,0)+\boxed{(-1)}\cdot (0,1,-1)+\boxed{0}\cdot (1,1,1) \ , \\ \\ (-3,1,-1) &=\boxed{-6}\cdot (1,1,0)+\boxed{4}\cdot (0,1,-1)+\boxed{3}\cdot (1,1,1) \ , \\ \\ (2,-1,2) &=\boxed{3}\cdot (1,1,0)+\boxed{(-3)}\cdot (0,1,-1)+\boxed{-1}\cdot (1,1,1) \ . \end{aligned} $$


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