A website that has programs that will do most operations in this course (an online calculator for matrices) of a Subspace This is a hypercube

Dimension - Hypercubes

Dimension A basis of a subspace consists of the fewest number of vectors needed to span the subspace. The basis of a subspace V in R n consist of the same number of vectors. The number of vectors in the basis of a subspace V is called the dimension of V.

Standard basis The vectors Form as standard basis for R n

How to find a basis for the image Find all of the columns that are independent. There are 2 ways this is generally done. 1)If the matrix is simple do by inspection. (use number sense to determine which columns are independent). 2) Put the matrix in reduced row echelon form. Every column that has a leading one in this form is independent. The columns (in the un-simplified form) form a basis for the image. Note: the number of vectors in the basis of the image is the dimension of the image

Recall this problem: Find the basis of the kernel and the image of Find the dimension of the image and the kernel.

Find the basis of the image by inspection Consider the columns 1)The first column is in the basis for the image because it is non-zero 2)The second column is not in the basis for the image because it is a multiple of column 1 3) The third column is not in the basis because it is the zero vector 4)The fourth column is in the basis because it is not a multiple of column 1 (the only vector that is in our basis so far) 5)The 5 th column is not in the basis because it is the sum of column 1 and 4 Answer: columns 1 and 4 form a basis of the image The dimension of the image is 2

Find the basis of the kernel Set Ax = 0 Solve and find all solutions Find rref(A) x 2, x 3 and x 5 are free variables Set x 2 = r, x 3 = s, x 5 = t x 1 = -2r –t x 2 = r x 3 = s x 4 = -t x 5 = t r s The three column vectors are a basis for the kernel The dimension of the kernel is 3

Problem 6 Find the basis of the kernel (null space) and image (column Space) of the matrix

Problem 6 Solution

Problem 10 Find the basis of the kernel (null space) and image (column Space) of the matrix Solve by inspection

Problem 10 Solution

What is the relationship between the dimension of the image and the dimension of the kernel?

The Fundamental Theorem of Linear Algebra For an nxm matrix Dim(Im(A)) + Dim (Ker(A)) = m For an nxn matrix Of course for a square matrix the dim(Im(A)) + Dim(ker(A)) = n Why is this true?

Problem 16 Find the basis of the kernel (null space) and image (column Space) of the matrix

Problem 16 Find the basis of the kernel (null space) and image (column Space) of the matrix Columns 1,2 and 4 appear to be independent, Hence they form a basis of the image. Due to the fundamental thm of LA we need To find only 1 vector to form a basis of the Kernel. Find 1 vector that results in zero

For an nxn matrix A, the following statements are equivalent 1 A is invertible 2. The linear system Ax=b has a unique solution 3. rref(A) = I 4. The rank of (A) = n 5. Im(A) = R n 6.Ker(A) = 0 7. The column vectors of A form a basis of R n 8. The columns vectors of A span R n 9. The column vectors of A are linearly independent. 10. Det A ≠ 0

Problem 29

29 Solution The dimension of this subspace is 2

Problem 39

Problem 39 Solution

Problem 34

Problem 34 Solution

I found a new technique for working in higher dimensions… When ever I run into a hypercube I give it Ritalin - Mr. Whitehead A Mathematician (M) and an Engineer (E) attend a lecture by a Physicist. The topic concerns Kulza-Klein theories involving physical processes that occur in spaces with dimensions of 9, 12 and even higher dimensions. The M is sitting, clearly enjoying the lecture, while the E is frowning and looking generally confused and puzzled. By the end the E has a terrible headache. At the end, the M comments about the wonderful lecture. E: "How do you understand this stuff?" M: "I just visualize the process" E: "How can you POSSIBLY visualize something that occurs in 9- dimensional space?" M: "Easy, first visualize it in N-dimensional space, then let N go to 9 Homework p odd, 37