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# Matrix algebra.....there is no spoon watch

1. Hi everyone,

I just finished my first year of Mathematics at Bath University and failed a unit on matrices rather badly

So I have to basically redo the exam paper, and it's confusing me a bit! If anyone from Bath who's done the paper well and can remember anything about it give me some pointers with some of the questions that would be great!

Here are the questions I'm stuck on:

4) Consider the following linear system in :

Calculating the rank of the augmented matrix and using the main theorem on linear systems, explain the geometric sense of the set of all the solutions of the given system.

6) Let , and be three mutually distinct eigenvectors of an x matrix A and let v1, v2 and v3 be the corresponding eigenvectors. Prove that is a set of linearly independent vectors.

8) Prove that the transposed rows of an n x n real orthogonal matrix Q form an orthonomal basis in .

9) Show that a real n x n matrix A is diagonalisable if the real eigenvectors of A span .

Any help would be much appreciated

Thanks
2. 4) i have no idea what the augmented matrix is. i've looked it up on wikipedia, but i don't see what it's meant to be referring to in this question?? anyway, the set of solutions forms an (n-1) dimensional plane, normal (1,1,...,1). we can write the equation as x.(1,1,...,1) = 1, then it's rather obvious. then, this 'rank' you talk of is (n-1), since, put simply, we can choose x_1, ..., x_(n-1) as we like (then forcing x_n to be 1 - x_1 - ... - x_(n-1))

6) suppose otherwise. apply the matrix to some expression of one e.vec as a linear combination of the other. derive a contradiction.

8) what's the definition of an orthogonal matrix?? write Q^T = (a b c) where a,b,c are column vectors (the things we want to show formal an orthonormal basis)

9) well, let's find the transformation matrix. its column (?) vectors will be the e.vecs of A, and since they span R^n, the inverse exists. hence we can diagonalise it.
3. (Original post by Chewwy)
then it's rather obvious. then, this 'rank' you talk of is (n-1), since, put simply, we can choose x_1, ..., x_(n-1) as we like (then forcing x_n to be 1 - x_1 - ... - x_(n-1))
I am probably mistaken, but are you sure? (See spoiler)

Spoiler:
Show
Some theorem says that, for a matrix A, rank(A) = rowrank(A) = columnrank(A). Rowrank is the number of linearly independent vectors in the rows of the matrix, and similarly for columnrank. Clearly there is only one row for the matrix, so we would expect the rank to be 1. Checking this with the columns, we see that each of the columns, we have (n+1) elements, but all the elements are linear combinations of each other, because of the equation given to us. So the columnrank is 1. Either way, we find the rank to be 1?
4. (Original post by Chewwy)
4) i have no idea what the augmented matrix is
its nothing special its just a matrix with a line between the columns for example solving linear systems;

2x+1y = 3
x-2y = 7

(2 1 | 3)
(1 -2 | 7)
5. (Original post by Kolya)
I am probably mistaken, but are you sure? (See spoiler)

Spoiler:
Show
Some theorem says that, for a matrix A, rank(A) = rowrank(A) = columnrank(A). Rowrank is the number of linearly independent vectors in the rows of the matrix, and similarly for columnrank. Clearly there is only one row for the matrix, so we would expect the rank to be 1. Checking this with the columns, we see that each of the columns, we have (n+1) elements, but all the elements are linear combinations of each other, because of the equation given to us. So the columnrank is 1. Either way, we find the rank to be 1?
ok then. bit of a funny thing to find though. i suppose 'the main theorem on linear systems' would explain everything..

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