§ 3

All square matrices of order 2 are multiplied on the right by the matrix

This defines a map $\mathrm{A}$ of the space of matrices of order 2 into the space of matrices of sizes 2 × 3.

What condition must the matrix $C$ of sizes 2 × 3 satisfy in order for the map $\mathrm{A}$, defined in problem 1, to be injective? Can it be surjective?

Let $\mathcal{C}^{k}$ be the linear space of functions having $k>1$ continuous derivatives on the segment $[0,1]$. Differentiation maps $\mathcal{C}^{k}$ into $\mathcal{C}^{k-1}$. Verify that this is a linear map. Will it be

Let $\mathrm{A}: \mathcal{L} \rightarrow \overline{\mathcal{L}}$ and $\mathcal{M}=\mathrm{A}(\mathcal{L})$. Define a map $\mathrm{A}^{\prime }: \mathcal{L} \rightarrow \mathcal{M}$ by the equality $\mathrm{A}^{\prime }(x)=\mathrm{A}(x)$. Prove that

Let $\mathcal{L}=\mathcal{L}_{1} \oplus \mathcal{L}_{2}$ and $x=x_{1}+x_{2}, x_{1} \in \mathcal{L}_{1}, x_{2} \in \mathcal{L}_{2}$. Define transformations $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ of the space $\mathcal{L}$ by the formulas: $\mathrm{P}_{1}(x)=x_{1}$ and $\mathrm{P}_{2}(x)=x_{2}$ (such transformations are called projections). Prove that

where $\mathrm{O}$ is the zero transformation and $\mathrm{E}$ is the identity transformation.

For any linear map $\mathrm{A}: \mathcal{L} \rightarrow \overline{\mathcal{L}}$ there exist bases in $\mathcal{L}$ and $\overline{\mathcal{L}}$ such that $A$ will have the matrix

( $E_{r}$ is the identity submatrix of order $r$, the remaining entries, if there are any, equal zero.) Prove this theorem by reducing the matrix of the linear map to the form (4) by elementary row and column transformations.

Let $\mathrm{A}$ be a linear map. Is it true that

Prove that the rank of a product of maps does not exceed the ranks of these maps. (This will be another proof of the estimate for the rank of a product of matrices from § 3 of ch. V: the rank of a product of two matrices does not exceed the ranks of the factors, $\mathrm{Rg} A B \leqslant \mathrm{Rg} A$ and $\mathrm{Rg} A B \leqslant \mathrm{Rg} B$.)

Let $\mathrm{B}$ be injective. Prove that for any $\mathrm{A}$ one has $\mathrm{RgBA}=\mathrm{RgA}$.