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Rodrigo de Azevedo
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How to find the kernel and image of a linear mapping?

Given the linear transformation:

$T_3 : \mathbb{C}^3 \to \mathbb{C}^3 , (x_1, x_2, x_3) \mapsto (x_1+x_2, x_1+x_3, x_1 - x_2+2x_3)$$$T_3 : \mathbb{C}^3 \to \mathbb{C}^3 , (x_1, x_2, x_3) \mapsto (x_1+x_2, x_1+x_3, x_1 - x_2+2x_3)$$

I need to find the image and kernel, and basis for both of them. I know that the kernel is the set of vectors for which the linear transformation will map to zero, and this can be found by setting up a system of equations like such:

$x_1 + x_2 = x_1 + x_3 = x_1 - x_2 +2x_3 = 0$$$x_1 + x_2 = x_1 + x_3 = x_1 - x_2 +2x_3 = 0$$

But once here, how would I format my answer for $ker(T_3)$, and how would I then find a basis for that? Secondly I understand the image of a linear transformation to be the span of the target vectors from the linear transformation, so would the $im(T_3)$ just be $\mathbb{C}^3$ ?

Thank you!

How to find the kernel and image of a linear mapping

Given the linear transformation:

$T_3 : \mathbb{C}^3 \to \mathbb{C}^3 , (x_1, x_2, x_3) \mapsto (x_1+x_2, x_1+x_3, x_1 - x_2+2x_3)$

I need to find the image and kernel, and basis for both of them. I know that the kernel is the set of vectors for which the linear transformation will map to zero, and this can be found by setting up a system of equations like such:

$x_1 + x_2 = x_1 + x_3 = x_1 - x_2 +2x_3 = 0$

But once here, how would I format my answer for $ker(T_3)$, and how would I then find a basis for that? Secondly I understand the image of a linear transformation to be the span of the target vectors from the linear transformation, so would the $im(T_3)$ just be $\mathbb{C}^3$ ?

Thank you!

How to find the kernel and image of a linear mapping?

Given the linear transformation:

$$T_3 : \mathbb{C}^3 \to \mathbb{C}^3 , (x_1, x_2, x_3) \mapsto (x_1+x_2, x_1+x_3, x_1 - x_2+2x_3)$$

I need to find the image and kernel, and basis for both of them. I know that the kernel is the set of vectors for which the linear transformation will map to zero, and this can be found by setting up a system of equations like such:

$$x_1 + x_2 = x_1 + x_3 = x_1 - x_2 +2x_3 = 0$$

But once here, how would I format my answer for $ker(T_3)$, and how would I then find a basis for that? Secondly I understand the image of a linear transformation to be the span of the target vectors from the linear transformation, so would the $im(T_3)$ just be $\mathbb{C}^3$ ?

Thank you!

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Nipster
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How to find the kernel and image of a linear mapping

Given the linear transformation:

$T_3 : \mathbb{C}^3 \to \mathbb{C}^3 , (x_1, x_2, x_3) \mapsto (x_1+x_2, x_1+x_3, x_1 - x_2+2x_3)$

I need to find the image and kernel, and basis for both of them. I know that the kernel is the set of vectors for which the linear transformation will map to zero, and this can be found by setting up a system of equations like such:

$x_1 + x_2 = x_1 + x_3 = x_1 - x_2 +2x_3 = 0$

But once here, how would I format my answer for $ker(T_3)$, and how would I then find a basis for that? Secondly I understand the image of a linear transformation to be the span of the target vectors from the linear transformation, so would the $im(T_3)$ just be $\mathbb{C}^3$ ?

Thank you!