Recent content by tas3113

the input vector space is 5. so if the dim(image) is 3 - then the dim(kernel) must be 2?

so wouldn't that mean all the columns and rows are linearly dependent?

just to clear this up. any row/column with all zeros is linearly independent? in that case, all the rows are linearly independent. as for the columns, the 3rd and 4th columns are dependent.

\left|\begin{array}{ccccc}-2&8&-12&8&0\\0&-6&0&0&0\\-2&0&0&0&0\\0&0&0&0&0\end{array}\right| ok. so the dim(ker(A)) = 3 and dim(image(A)) = 2

the last column is all zeros, so that is linearly dependent. Rows 3 and 4 are linearly dependent too, as with row 5. so that leads me back to dim(image(A)) = 3 and dim(ker(A)) = 2, which is what i had before but that was wrong?

\begin{align*}T(x^2) &= 2f(x) - f(x-1) - f(x+1)\\ &= 2x^2 - (x^2 - 2x + 1) - (x^2 + 2x + 1) \\ &=-2 \end{align*} \left|\begin{array}{cccc}-2 & 0 & -2 & 0 \\ 8 & -6 & 0 & 0\\ -12 & 0 & 0 & 0\\8 & 0 & 0 &...

\begin{align*}T(x^4) &= 2f(x) - f(x-1) - f(x+1)\\ &= 2x^4 - (x^4 - 4x^3 + 6x^2 - 4x + 1) - (x^4 + 4x^3 + 6x^2 + 4x + 1) \\ &= 8x^3 -12x^2 + 8x - 2 \end{align*} \begin{align*}T(x^3) &= 2f(x) - f(x-1) - f(x+1)\\ &= 2x^3 - (x^3 - 3x^2 + 3x - 1) - (x^3 +...

i've been searching through the book trying to find something on linear transformations similar to this but couldn't find anything. i have no idea what to do with f(x) --> 2f(x) - f(x-1) - f(x+1). are you saying to like plug each element of x^4,x^3,x^2,x^1,1 one by one...

ok. and likewise for one whose kernel is two-dimensional would simply be a 5 x 5 matrix with 3 linearly independent columns. its all slowly starting to make sense. as for the other question: 2. Find the matrix representation in the standard basis of the linear transformation from P4 to P3...

dim(range(M)) = 4 since it spans R^4. Thus, dim(null(M)) = 1 D = 5 D = n. The number of columns, since the range and null have to add up to the number of columns. It would be for one-dimension?

\left|\begin{array}{ccccc}1&1&1&1&1 \\ 2&3&1&3&2\\1&1&1&3&1\\1&4&1&4&1\\5&5&5&5&5\end{array}\right| thats an example. the first four are linearly independent and the first and last are dependent. were you asking for a more generic answer though? like with variables? as for dim(null M) and...

it would be a 5 x 5 matrix.

ok. I am going to read this through a few more times to understand it. ill make sure to ask if i have any questions. thanks for your help on this problem

So T is also a vector? in this case a 1 x 5 matrix. so: T = {0, 0, 0, 0, 0} for dimension of 0 T = {0, 0, 0, 0, a} for dimension of 1 T = {0, 0, 0, b, a} for dimension of 2

You say apply the map to it for both, but I don't quite understand what that means exactly. This is my thought process so far. a_3 x^3 + a_2 x^2 + a_1 x + a_0 Is T what you are multiplying by to get the result? So for a one-dimensional kernel...