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$$B = \left(\begin{matrix} 0.4 & 0 & 0 & 0 \\ 0 & 0.4 & 0 & 0 \\ 0 & 0 & 0.4 & 0 \end{matrix}\right)$$

Here, $B \in \mathbb{C}^{3 \times 4}$ where $\mathbb{C}$ is complex field. Let $T = \mathbb{C}^3$ and $e = (0,0,0,1)^t \in \mathbb{C}^4$ and $S = \mathrm{span}(e)$. $T$ and $S$ are vector spaces.

How to show that $\mathrm{range}(B) \subset T$ and can we conclude the similar relation between the null space of $B$ and $S$?

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i tried to find rank of given matrix B which is 3. i am confused why these two spaces are not equal despite of having same number of rank? – srijan May 7 '12 at 13:05
also i found null space of B that is coming out to be span{0,0,0,c}where c is constant. so can i conclude that null (B)= S – srijan May 7 '12 at 13:06
Which two spaces are not equal? Why do you say they are not equal? What is your problem with the nullspace? – Gerry Myerson May 7 '12 at 13:10
Note that $X\subset Y$ does not exclude the possibility that $X=Y$, if that's what you're confused about. – Daan Michiels May 7 '12 at 13:11
The inclusion is not proper. For most authors, $\text{range}(B)\subset T$ means the same thing as $\text{range}(B)\subseteq T$, i.e. that $\text{range}(B)$ is either a proper subset of $T$, or that $\text{range}(B)=T$. In this case, $\text{range}(B)=T$. – Daan Michiels May 7 '12 at 13:37

Just to get you thinking about this problem. Note that $B$ is a $3\times 4$ matrix. So as a linear map we can view $B$ as a map from $\mathbb{C}^4 \to \mathbb{C}^3$, so you already have that the range of $B$ is contained in $ T = \mathbb{C}^3$.

About the nullspace of $B$. Consider the equation

$$\begin{pmatrix} 0.4 & 0 & 0 & 0 \\ 0 & 0.4 & 0 & 0 \\ 0 & 0 & 0.4 & 0 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.$$

The solution to this equation is clearly that $x_1 = x_2 = x_3 = 0$ and $x_4 \in \mathbb{C}$. Hence the nullspace is isomorphic (as a vector space) to $\mathbb{C}$ which is isomorphic to the span of the element $e$ in your question.

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