Let $\phi: $$\Bbb{C^3}\to\Bbb{C^2}$ be linear transformation such as $$\phi(x_1,x_2,x_3)=(x_1+x_2,ix_3) $$ Then base of $\ker\phi=(-x_2,x_2,0) $ but what is $\dim \ker \phi?$
base of image $\phi $ =?
Let $\phi: $ $\Bbb{C^3}\to\Bbb{C^2}$ be linear transformation
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linear-algebra
linear-transformations
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0@TheMeff: Do you know how to (i) Express this type of linear transformation as a matrix, (ii) Row reduce the transformation matrix to echelon form, (iii) Use the (reduced) row-echelon form to read out bases of the null space (kernel) and column space (image)? – 2017-01-28
1 Answers
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The dimension of a vector space is the number of elements in a basis of the vector space. So as you have already found a basis it should be easy to find the dimension.
For the second question note that $\phi(1,0,0) = (1,0)$ and $\phi(0,0,-i) = (0,1)$. Therefore we have that $\mathbb{C}^2 = \langle(1,0),(0,1)\rangle \subseteq \text{Im}(\phi) $. Therefore $\text{Im}(\phi) = \mathbb{C}^2$ and $\{(1,0),(0,1)\}$ is a basis of $\text{Im}(\phi)$.