Linear Transformation $\operatorname{Im} f \neq \operatorname{Im} (f \circ f)$

Question

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 $f(x,y,z) = (x+z,2x+y,3x+y+z)$ for all $x, y, z \in \mathbb{R}$. $(1)$ Find the matrix representation of $f$ and $f \circ f$ with respect to the standard basis of $\mathbb{R}^{3}$. $(2)$ Find a base and the dimension of the vector space $\operatorname{Im} f$. $(3)$ Explain why $\operatorname{Im} f \neq \operatorname{Im} (f \circ f)$.

Since I'm having difficulties for $(3)$ I'm going to share my work for $(1)$ and $(2)$.

$(1)$ We consider the standard basis $\mathcal{B}=\{(1,0,0), (0,1,0), (0,0,1)\}$ 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

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

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

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

Now, since $f$ is represented by A, then $f \circ f$ is represented in $\mathcal{B}$ by the matrix $A⋅A$. One has

$$A^{2} = \begin{pmatrix}4 & 1 & 2\\4 & 1 & 2\\8 & 2 & 4\end{pmatrix}.$$

$(2)$ Since the set $\mathcal{B}=\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}$, as a base of $\mathbb{R^{3}}$, spans $\mathbb{R^{3}}$, it follows that the set $f(\mathcal{B})=\{f(1, 0, 0), f(0, 1, 0), f(0, 0, 1)\}$ spans $\operatorname{Im} f$. We have

$$\operatorname{Im} f = \operatorname{Span}\{f(1, 0, 0), f(0, 1, 0), f(0, 0, 1)\} = \operatorname{Span}\{(1, 2, 3), (0, 1, 1), (1, 0, 1)\}$$ and

$$\begin{pmatrix}1 & 2 & 3\\0 & 1 & 1\\1 & 0 & 1\end{pmatrix} \rightarrow \ldots \rightarrow \begin{pmatrix}1 & 0 & 1\\0 & 1 & 1\\0 & 0 & 0\end{pmatrix}.$$

Therefore $\operatorname{Im} f = \operatorname{Span}\{(1,0, 1), (0, 1, 1)\}$ and $\dim (\operatorname{Im} f) = 2$.

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Is my work correct for $(1)$, $(2)$? Any help for $(3)$ would be appreciated. At first, I wanted to say that since the matrix representation of $f$ and $f \circ f$ are different, then $\operatorname{Im} f \neq \operatorname{Im} (f \circ f)$ but I don't think this is a good explanation (I don't even know if this is true in general).

Answer 1

Inspect the def'n of $f.$ Notice that if $f(x,y,z)=(u,v,w)$ then $w=u+v.$ So if $z\in im(f)$ then $z=(u,v,u+v)$ for some $u, v.$

Applying the def'n of $f$ we have $f(u,v,u+v)=(r,r,2r)$ where $r=2u+v.$ So $im (f^2)\subset \{r(1,1,2):r\in \mathbb R\}$, implying $dim (im(f^2))\leq 1.$ (In fact it is equal to $1$ .)

You have already shown that $dim(im (f))=2.$ Therefore $im(f)\ne im (f^2).$

Answer 2

HINT: $$\dim (\text{Im}(f \circ f)) = 1$$

Answer 3

We have $A^{2} = \begin{pmatrix}4 & 1 & 2\\4 & 1 & 2\\8 & 2 & 4\end{pmatrix}$.

From this representation it is easy to see that

$\operatorname{Im} (f \circ f) = \operatorname{Span}\{(1,1, 2)\}$

Answer 4

At first, I wanted to say that since the matrix representation of $f$ and $f \circ f$ are different, then $\operatorname{Im} f \neq \operatorname{Im} (f \circ f)$ but I don't think this is a good explanation (I don't even know if this is true in general).

Not, that won't stick. There are lots of different matrices with identical images -- for example, all invertible $n\times n$ matrices have the same image. Even in one dimension, your argument wouldn't work for $f(x)=2x$ -- the matrix representations of $f$ and $f\circ f$ would be $[2]$ and $[4]$, but the image of each is all of $\mathbb R$.

"Explain" is a bit of a fuzzy word, but if you fiddle around with the eigenspaces a bit, you should be able to find a basis in which the matrix of $f$ is $$ \begin{pmatrix} 3 & 0 & 0 \\ 0 & 0 & 1 \\ 0& 0& 0 \end{pmatrix} $$ where it is a bit easier to see what's going on.