- #1

Kernul

- 211

- 7

## Homework Statement

Being f : ℝ

^{4}→ ℝ

^{4}the endomorphism defined by:

ƒ((x, y, z, t)) = (3x + 10z, 2y - 6z - 2t, 0, -y+3z+t)

Determine the base and dimension of Im(ƒ) and Ker(ƒ). Complete the base you chose in Im(ƒ) into a base of R

_{4}.

## Homework Equations

Matrix A:

$$\begin {bmatrix}

3 & 0 & 10 & 0\\

0 & 2 & -6 & -2\\

0 & 0 & 0 & 0\\

0 & 1 & 2 & 1

\end{bmatrix}$$

dim(Im(ƒ)) = rank(A)

dim(Ker(ƒ)) = number of columns - rank(A)

## The Attempt at a Solution

So, first I have to know the rank of the matrix. The fourth row is actually the second row divided by -2. So:

A

_{4}= (-1/2) * A

_{2}

At this point I know that the matrix has a rank of 3, but there is a null row. Does this mean that the rank is 2? When a row(or multiple row) is 0 the rank is the number of the remaining non-null rows?

This is no the only question so I'll continue by assuming it's a rank of 2, so:

rank(A) = 2 = dim(Im(ƒ))

At this point while I look at other exercises done by my professor I see that she write like this for the base of Im(ƒ):

B

_{Im(ƒ)}= [A

^{1}, A

^{2}, A

^{3}, ...] (This if it was my exercise: B

_{Im(ƒ)}= [A

^{1}, A

^{2}])

I want to know why this is the base because I really don't understand.

At this point we have to determine the dimension and the base of Ker(ƒ).

The dimension would be:

dim(Ker(ƒ)) = 4 - 2 = 2

To determine the base of Ker(ƒ), from what I understood, you have to take the rows and set 2(because the dimension of Ker(ƒ) is 2) of x, y, z and t, equal to a scalar(a and b), put them into a system and then solve it until you get the two columns of the base of Ker(ƒ). So it would be something like this:

{3x +10z = 0

{2y - 6z - 2t = 0

{z = b

{t = a

Solving it we have:

{x = (-10/3)b

{y= 3a + b

{z = b

{t = a

And this:

Ker(ƒ) = {$$\begin {bmatrix}

0\\

3\\

0\\

1

\end{bmatrix}$$ * a +

$$\begin {bmatrix}

-10/3\\

1\\

1\\

0

\end{bmatrix}$$ * b : a, b ∈ ℝ}

And the base of Ker(ƒ) would be:

B

_{Ker(ƒ)}= [$$\begin {bmatrix}

0\\

3\\

0\\

1

\end{bmatrix}$$,

$$\begin {bmatrix}

-10/3\\

1\\

1\\

0

\end{bmatrix}$$

But why? I don't understand why this it how to calculate the bases.

I'm so sorry for all the confusion about the matrix but I don't know how to make them better. If you know, could you please tell me?

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