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Let $T:R^4 \to R^4$ be linear transformation such that $T(x)=Ax$, when

$A = \begin{bmatrix}1&2&3&4\\1&a_1&b_1&c_1\\1&a_2&b_2&c_2\\1&a_3&b_3&c_3\end{bmatrix}$

Suppose $\dim(\mathrm{Im}(T)) < \dim(\ker(T))$, find ${a_i,b_i,c_i}$ when $i=1,2,3$, and find a basis and dimension for $\mathrm{Im}(T), \ker(T)$.

What I did: I figured that $\dim(\mathrm{Im}(T))=1$ so $rank(A)$ should be $1$ also, so thats means that there is only non-zero row in the matrix, right? but does it mean $a_i=2, b_i=3, c_i=4$ for $i=1,2,3$?

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1 Answer 1

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Since the first row and column are specified, $\dim(\mathrm{Im}(T)) \ge 1$ and $\dim(\ker(T)) \le 3$. Since the sum of these two dimensions must be 4, the only way to satisfy the inequality is as you said: the rank is 1. Since the first row is given, then every other row must be some multiple of the first. Since every other rows starts with 1, then all the rows must be the same.

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