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Suppose we have a linear map $T: \mathbb{R}^4 \to \mathbb{R}^4$ such that $\operatorname{dim} (\operatorname{Im}T) < \operatorname{dim} (\operatorname{ker}T)$ and the matrix representing $T$ in the basis $B=((1,1,1,1),(1,1,1,0),(1,1,0,0),(1,0,0,0))$:

$$[T]_B = \begin{pmatrix} 1 & 2 & 3 & 4\\ 1 & a_1 & b_1 & c_1\\ 1 & a_2 & b_2 & c_2\\ 1 & a_3 & b_3 & c_3 \end{pmatrix}$$

I need to find the numbers $a_i, b_i, c_i ~~~ (1 \leq i \leq 3)$

My attempt: from $\operatorname{dim} (\operatorname{Im}T) < \operatorname{dim} (\operatorname{ker}T)$ I conclude that $\operatorname{dim} (\operatorname{Im}T)=1$, $\operatorname{dim} (\operatorname{ker}T)=3$ (because $\operatorname{dim} (\mathbb{R}^4)=4, T(1,1,1,1) \neq 0 \Rightarrow \operatorname{Im}T \neq \{0\}$ and the rest follows from rank–nullity theorem).

My problem is with the unknown scalars. Because $\operatorname{dim} (\operatorname{ker}T)>0$ the determinant of $T$ must be equal to zero ($\det{T}=0$), because only then $[T]_B \vec{v} = \vec{0}$ will have non-trivial solutions. However the expression for $\det{T}$ is horrific and very complex, even for the $4 \times 4$ matrix. For sure I'm missing something simple yet crucial. I don't ask for a solution - just a clue or an insight. What am I missing?

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3 Answers 3

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Let denote $e_i,\;i=1,\ldots,4$ the vectors of the given basis. Since the dimension of the image is $1$ then the image of $T$ is a subspace of $\Bbb R^4$ spanned by one vector and since $T(e_1)=(1,1,1,1)^T=v=e_1+e_2+e_3+e_4$ then $$\operatorname{Im}(T)=\operatorname{span}(v)$$ hence $$T(e_2)=(2,a_1,a_2,a_3)^T=2e_1+a_1e_2+a_2e_3+a_3e_4=kv$$ Can you take it from here?

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  • $\begingroup$ I think it should be $[T(e_1)]_B=(1,1,1,1)^T$ (by mistake you equate the coordinates vector of the image to the image itself). I'll try to continue from here. Thanks. $\endgroup$
    – Ferl
    Jan 9, 2015 at 20:39
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Note that $\dim\operatorname{Im}T = \dim\operatorname{Col}[T]_{B} = \dim\operatorname{span}\{v_1, v_2, v_3, v_4\}$ where $v_i$ is the $i^{\text{th}}$ column vector of $[T]_{B}$. If $\dim\operatorname{Im}T = 1$, what can you say about the vectors $v_2$, $v_3$, $v_4$ in relation to $v_1$?

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  • $\begingroup$ That the other vectors are proportional to $v_1$. That is, $v_1$ can be a basis of $\operatorname{span}\{v_1, v_2, v_3, v_4\}$. $\endgroup$
    – Ferl
    Jan 9, 2015 at 20:34
  • $\begingroup$ That's correct. So $v_2 = av_1$ for some $a \in \mathbb{R}$. Comparing the first components of both sides will allow you to determine $a$, and hence $v_2$. Likewise for $v_3$ and $v_4$. $\endgroup$ Jan 9, 2015 at 20:48
  • $\begingroup$ thank you very much. Simple indeed. $\endgroup$
    – Ferl
    Jan 9, 2015 at 20:50
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    $\begingroup$ by the way, where can I see a proof that $\dim\operatorname{Im}T = \dim\operatorname{Col}[T]_{B}$? I know that in general $\operatorname{Im}T=\operatorname{span}(T(e_1),...,T(e_n))$ (where $e_i$ some basis vector). But I'm not sure how to prove that the coordinate vectors of the images are also spanning $\operatorname{Im}T$. $\endgroup$
    – Ferl
    Jan 10, 2015 at 9:38
  • $\begingroup$ is it because the isomorphism between the $n$-dimensional vector space $V$ and $\mathbb{R}^n$? $\endgroup$
    – Ferl
    Jan 10, 2015 at 16:56
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any matrix representing rank one matrix is of the form $ab^T.$ in the question you have $a = \pmatrix{1\\1\\1\\1}$ and $b^T = \pmatrix{1&2&3&4}.$ so that gives $$ a_1 = 2, b_1 = 3 \text{ and } c_1 =4$$

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