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I'm learning linear algebra, specifically linear transformations, and need help with the following exercise:

Consider the endomorphism $f: \mathbb{R^{3}} \to \mathbb{R^{3}}$ defined by $$f(x,y,z) = (x+2y+3z,-x+4z,2y+2z)$$ for all $x, y, z \in \mathbb{R}$. Show that $f$ is invertible and find $f^{-1}$.

My work and thoughts:

We consider the standard basis $\mathscr{B}=\{(1,0,0), (0,1,0), (0,0,1)\}$ of $\mathbb{R^{3}}$. Given that $f$ is a linear transformation, it can be represented by a matrix $A$ whose columns are the images of the standard basis vectors. We have

$$f(1, 0, 0) = (1, -1, 0), \quad f(0, 1, 0) = (2, 0, 2), \quad f(0, 0, 1) = (3, 4, 2).$$

Hence, the matrix representation of $f$ with respect to the basis $\mathscr{B}$ is given by

$$A = \begin{pmatrix}1 & 2 & 3\\-1 & 0 & 4\\0 & 2 & 2\end{pmatrix}.$$

From here, I'm not sure what I should actually do. Do I need to show that the matrix $A$ is invertible? If $A$ is invertible, does this mean that $f$ is also invertible and, more importantly, why? If so, how do I explicitly write $f^{-1}$?

To save me some time, I have evaluate the inverse of $A$ in Mathematica and found

$$A^{-1} = \begin{pmatrix}\frac{4}{5} & -\frac{1}{5} & -\frac{4}{5}\\-\frac{1}{5} & -\frac{1}{5} & \frac{7}{10}\\\frac{1}{5} & \frac{1}{5} & -\frac{1}{5}\end{pmatrix}.$$

3 Answers 3

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The matrix $A$ representing a mapping $f$ is invertible if and only if the mapping $A$ is invertible.

This fact should be (and, don't worry, it will be) intuitively true for you, and in fact, if you don't yet see why this would be true, then you need more practice on the subject. It comes with time and work, so don't worry too much.

It's most important to know that multiplying matrices is effectively the same thing as calculating compositions of the mappings that they represent. So, if $f$ is represented by $A$, and $g$ by $B$ (both in basis $\mathcal B$), then $f\circ g$ is represented (in $\mathcal B$) by the matrix $A\cdot B$.

Once you internalize this fact (that, in some way, $A$ and $f$ are the same thing), it's clear that

  • if $f$ is invertible, then the matrix representing the mapping $f^{-1}$ must be the matrix inverse of $A$
  • if $A$ is invertible, then $A^{-1}$ is the representation of a mapping that must be the inverse of $f$.

Another way of looking at this is that in order for $f$ to be invertible, there must exist, for each $a,b,c$, exactly one value $(x,y,z)$ such that $f(x,y,z)=(a,b,c)$.

This leads you to write down the set of equations, and realize that the equations you are solving are

$$\begin{pmatrix}1 & 2 & 3\\-1 & 0 & 4\\0 & 2 & 2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}a\\b\\c\end{pmatrix}$$

and this system of equations has a unique solution for all possible values $a,b,c$ if and only if the matrix is invertible.

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$f$ is the left multiplication of the coordinate vector $\begin{pmatrix}x\\y\\z\end{pmatrix}$ by the matrix $A$, because:

$$f\left(\begin{pmatrix}x\\y\\z\end{pmatrix}\right)=\begin{pmatrix}x+2y+3z\\-x+4z\\2y+2z\end{pmatrix} = \begin{pmatrix}1 & 2 & 3\\-1 & 0 & 4\\0 & 2 & 2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}$$

We can consider $g$ that is the left multiplication by the matrix $A^{-1}$. Then $$g\circ f\left(\begin{pmatrix}x\\y\\z\end{pmatrix}\right)=A^{-1}(A\begin{pmatrix}x\\y\\z\end{pmatrix})=(A^{-1}A)\begin{pmatrix}x\\y\\z\end{pmatrix}=I\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x\\y\\z\end{pmatrix}$$

Thus $g\circ f=id$. The same argument shows that $f\circ g=id$, because $AA^{-1}=I$.

You now only need to write $g$ in coordinate form. For you can just multiply $A^{-1}$ by $\begin{pmatrix}x\\y\\z\end{pmatrix}$.

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With your matrix $A$ we have

$f(x,y,z)=A*(x,y,z)^T$ for all $(x,y,z) \in \mathbb R^3$.

$f$ is invertible iff $ker(f)=\{0\}$

and

$A$ is invertible iff $ker(A)=\{0\}$.

Now show that $ker(f)=\{0\}= ker(A)$.

It results: $f$ is invertible iff $A$ is invertible.

Next put $g(x,y,z)=A^{-1}*(x,y,z)^T$ for $(x,y,z) \in \mathbb R^3$ and verify

$f(g(,x,y,z))=(x,y,z)=g(f(x,y,z))$ for all $(x,y,z) \in \mathbb R^3$.

Hence $f^{-1}=g$