Translating an Italian exercise precisely

I am trying to help a friend with his algebra course. However, his exercises are in Italian, and unfortunately he translates them poorly for me since he does not know the mathematical terms in English. I have tried to translate it myself, but without luck. I still do not know what exactly to do.

More precisely, it is the description in the exercise and exercise (a) that I do not understand completely; the rest I understand.

E2) Sia $$f\colon\mathbb{C}^4\to\mathbb{C}^4$$ una trasformazione lineare e si supponga che la matrice associata a $$f$$ rispetto alla base $$\mathcal{B} = \{\mathbf{e}_2; \mathbf{e}_1; \mathbf{e}_3+\mathbf{e}_4; \mathbf{e}_3+\mathbf{e}_2\}$$ su dominio e codominio ($$\mathbf{e}_i$$ sono i vettori della base canonica di $$\mathbb{C}^4$$) sia

$$\mathbf{A} = \begin{bmatrix}3&3&2&2\\ 3&3&2&2\\ 0&0&1&1\\ 0&0&1&1\end{bmatrix}$$

(a) Si determini la matrice $$\mathbf{B}$$ associata a $$f$$ rispetto alle basi canoniche.

(b) Si calcoli la dimensione dell'immagine di $$f$$.

(c) Si dica se la matrice $$\mathbf{B}$$ รจ diagonalizzabile.

(d) Si calcoli una base dello spazio nullo dell'applicazione lineare $$f$$.

• @Husky653 matrix $A$ represents the mapping when vectors are represented with respect to the specific base instead of the natural base. You have to find the base transformation matrix $M$ that turns vectors expressed with respect to this specific base into vectors expressed with respect to the natural base: $$x_n=Mx \Leftrightarrow x=M^{-1}x_n$$ then perform the transformation into the natural base of the linear map: $$y=Ax \Leftrightarrow My=MAx \Leftrightarrow y_n=MAM^{-1}x_n$$ that is $$B = MAM^{-1}$$ – polettix Feb 20 '16 at 19:19
• @PålGD that would be a different question I suppose? – polettix Feb 20 '16 at 19:20
• I take it you don't speak Italian, and your friend doesn't know mathematical terms in English. That combination should make helping him great fun. – Eckhard Feb 20 '16 at 23:22

Suppose $$f$$ is a linear transformation and suppose further that the matrix of $$f$$ with respect to the basis $$B = \{ e_2; e_1; e_3 + e_4; e_3 + e_2 \}$$ is the matrix $$A = ...$$, where the basis $$B$$ is used as the basis of both the domain and codomain ($$e_i$$ being the vectors of the standard basis of $$\mathbb{C}^4$$)

a. Determine the matrix for the transformation $$f$$ with respect to the canonical (standard) basis.

b. Calculate the dimension of the image of $$f$$ (i.e., the "rank" of $$f$$).

c. Say whether the matrix $$B$$ for $$f$$ is diagonalizable.

d. Calculate a basis for the null-space (or kernel) of the linear transformation $$f$$.

There you go. That's a pretty solid translation of the exercise. Now, perhaps., you can confidently work through it.

• Thanks to the various folks who edited my translation to make it more literal. Since I don't actually speak Italian, I just went with "what would an English speaking linear algebra teacher ask, given the mathematical notation in the problem?" combined with cognates and a knowledge of enough French and Spanish to make sense of most Italian math stuff I encounter. :) – John Hughes Feb 20 '16 at 17:04
• Do you have a hint towards how I can solve a)? I am usually used to have the exercise written in a different form and I think that they way it is written here is confusing me, I am only having problems with exercise a). A few hints just, not the solutions thank you. – Husky653 Feb 20 '16 at 17:12
• Suppose in $R^2$ you have a vector $v$ whose coordinates are $1, 3$ in the basis $e_1 + 4e_2, e_1 - e_2$. Then you can multiply $v' = Mv$, where $M = \begin{bmatrix} 1 & 1 \\ 4 & -1\end{bmatrix}$ to get $v' = \begin{bmatrix} 4\\ 1\end{bmatrix}$, which are the coordinates of $v$ in the standard basis. (The columns of $M$ are the two given basis vectors!) What matrix $K$ would undo that operation (i.e., send standard vectors to funny-basis vectors)? If you have a transform whose matrix in the funny basis is $A$, what will $M A K$ do? Convert from standard to funny; do transform; convert back.! – John Hughes Feb 20 '16 at 18:36

Let $f:\mathbb{C}^4 \rightarrow \mathbb{C}^4$ be a linear map and suppose that the matrix associated with $f$ with respect to the basis $\mathcal{B}=\{ e_2,e_1,e_3 + e_4, e_3 + e_2\}$ (where $e_i$ are the canonical basis vectors of $\mathbb{C}^4$) is (the matrix $A$ written above in the exercise).

(a) Determine the matrix $B$ associated with $f$ with respect to the canonical basis of $\mathcal{C}^{4}$.

(b) Calculate the dimension of the image of $f$.

(c) Say whether the matrix $B$ is diagonalizable.

(d) Calculate a basis of the null space of the linear map $f$.

• Do you have a hint towards how I can solve a)? I am usually used to have the exercise written in a different form and I think that they way it is written here is confusing me, I am only having problems with exercise a). A few hints just, not the solutions thank you. – Husky653 Feb 20 '16 at 17:17
• You can apply the matrix of the change of basis and then solve a linear system. Otherwise you can determine the linear transformation f using the column of the matrix writing every action of f on the column vector in the given basis B. Then once you have the linear transformation you can write the matrix in the canonical basis – Salvatore Feb 20 '16 at 18:53