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- Thread starter Jerome
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- Mar 5, 2012

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Welcome to MHB, Jerome!Consider d map f:R^4 into R^2 defines by f(x,y,z,w)=(2x+y+z+w,x+z-w). find the image and the kernel, pls include explanations pls..

Do you have definitions handy for

If so, we can see how we can apply them.

Let's start with the image.

In my book the image is the set of all vectors that can be "reached" by f.

Can for instance (1,0) be "reached"?

Or put otherwise, can you find an (x,y,z,w) in $\mathbb R^4$ that has (1,0) as its image?

And how about (0,1)?

If both can be reached, the image is $\mathbb R^2$, since (1,0) and (0,1) "span" $\mathbb R^2$.

- Jan 29, 2012

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Since f maps (x, y, z, w) to (2x+y+z+w,x+z-w) so the kernel must have 2x+ y+ z+ w= 0, x+ z- w= 0. If we add the two equations the "w"s cancel and 3x+ y+ 2z= 0 so that y= -3x-2z. Clearly w= x+ z. So (x, y, z, w)= (x, -3x- 2z, z, x+ z)- (x, -3x, 0, x)+ (0, -2z, z, z)= x(1, -3, 0, 1)+ z(0, -2, 1, 1) which tells you immediately what the dimension of the kernel and a basis for the kernel is.

Notice that this satifies the "rank nullity" theorem: if a linear transformation is from a vector space of dimension n to a vector space of dimension m, the rank (dimension of the image) and nullity (dimension of the kernel) add to n.

Here, that is 2+ 2= 4.

- Feb 15, 2012

- 1,967

\(\displaystyle B_1 = \{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)\}\)

\(\displaystyle B_2 = \{(1,0),(0,1)\}\)

The 2x4 matrix I obtain relative to these two bases is:

\(\displaystyle \begin{bmatrix}2&1&1&1\\1&0&1&-1 \end{bmatrix}\)

It should be clear immediately that this matrix has rank 2 (the two rows are linearly independent), so the image of f is of dimension 2, and the ONLY 2-dimensional subspace of \(\displaystyle \Bbb R^2\) is, of course, \(\displaystyle \Bbb R^2\) itself.

This settles the image question (f is surjective), but only tells us (via the rank-nullity theorem) that the kernel has dimension 2. So we actually need to FIND a basis for the kernel (or: equivalently, the null space of our matrix above). How do we do that?

We need to solve the following HOMOGENEOUS system of equations:

2x + y + z + w = 0

x + z - w = 0

Putting our 2x4 matrix into rref form, we get the matrix:

\(\displaystyle \begin{bmatrix}1&0&1&-1\\0&1&-1&3 \end{bmatrix}\)

This is equivalent to the system:

x + z - w = 0

y - z + 3w = 0

If we pick z = 1, w = 0, we get:

x + 1 = 0

y - 1 = 0, leading to the vector (-1,1,1,0).

If we pick z = 0, w = 1, we get:

x - 1 = 0

y + 3 = 0, leading to the vector (1,-3,0,1). These two vectors are clearly linearly independent. Are they elements of the kernel? We check:

f(-1,1,1,0) = (-2+1+1+0,-1+1+0) = (0,0)

f(1,-3,0,1) = (2-3+0+1, 1+0-1) = (0,0)

so {(-1,1,1,0),(1,-3,0,1)} is a basis for the kernel.

Does this agree with Hall's answer? If so, we should be able to express both:

(1,-3,0,1) and (0,-2,1,1) as a linear combination of our two vectors (remember, bases are not UNIQUE). And indeed:

(1,-3,0,1) = (0)(-1,1,1,0) + (1)(1,-3,0,1)

(0,-2,1,1) = (1)(-1,1,1,0) + (1)(1,-3,0,1)