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The following $m$ x $n$ matrix, decompose the first standard basis vector $e_1 = w + z \in \mathbb{R^n}$, where $w \in$ rowspace(A) and $z\in kerA$. Verify your answer by expressing $w$ as a linear combination of the rows of $A$.

a.) $\pmatrix{1&-2&1\\2&-3&2}$

b.) $\pmatrix{1&-1&0&3\\2&1&3&3\\1&2&3&0}$

For a, I got the $kerA = \pmatrix{-1\\0\\1}$ and the $rowspaceA = \pmatrix{1\\-2\\1},\pmatrix{2\\-3\\2}$.

The answer is:

$z = \pmatrix{\frac{1}{2}\\0\\-\frac{1}{2}}, w = \pmatrix{\frac{1}{2}\\0\\\frac{1}{2}} = - \frac{3}{2} \pmatrix{1\\-2\\1} + \pmatrix{2\\-3\\2}$

How did they get that?

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1 Answer 1

up vote 2 down vote accepted

$e_1=\pmatrix{1\\0\\0}=z+w=c_1\pmatrix{-1\\0\\1}+c_2\pmatrix{1\\-2\\1}+c_3\pmatrix{2\\-3\\2}$ and solve the equations.

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