Hi guys,

I have worked out an answer to the following question, but it's stretching my understanding of Linear Algebra to the limit, soâ€¦ if anybody could have a look at it, and just tell me whether my reasoning is sound, it would be a great help...

Let v_1,â€¦,v_k,u,w be vectors in the linear space V.

It is given that the equation x_1Â·v_1+â€¦x_kÂ·v_k=u has a single solution and that the equation x_1Â·v_1+â€¦+x_kÂ·v_k=w has no solution.

a. Prove that the group {v_1,â€¦,v_k} is linearly independent.

b. Find the dimension of Sp{v_1,â€¦,v_k,w}.

My answer:

a.

According to the information provided, the equation

x_1Â·v_1+â€¦x_kÂ·v_k=u has a single solution.

That means that for the solution to hold, all variables in this equation must have a fixed value, with no free variables. If u is the zero vector, each one of the variables must take on the value 0 for the single solution to hold: it is a given that there is a single solution only, so in this case it must be the trivial solution, as the trivial solution always exists in a homogeneous system. If the trivial solution is the only solution, then v_1 to v_k is linearly independent.

b.

per definition {v_1,â€¦,v_k} spans Sp{v_1,â€¦,v_k}, and it was shown that {v_1,â€¦,v_k} is linearly independent, so {v_1,â€¦,v_k} is a basis with dimension k. Sp{v_1,â€¦,v_k,w} has an additional member, w, which according to the information given is not a linear combination of {v_1,â€¦,v_k} (given that the equation x_1Â·v_1+â€¦+x_kÂ·v_k=w has no solution). Therefore {v_1,â€¦,v_k,w} is a basis of Sp{v_1,â€¦,v_k,w}, which dimension k+1.

I have worked out an answer to the following question, but it's stretching my understanding of Linear Algebra to the limit, soâ€¦ if anybody could have a look at it, and just tell me whether my reasoning is sound, it would be a great help...

Let v_1,â€¦,v_k,u,w be vectors in the linear space V.

It is given that the equation x_1Â·v_1+â€¦x_kÂ·v_k=u has a single solution and that the equation x_1Â·v_1+â€¦+x_kÂ·v_k=w has no solution.

a. Prove that the group {v_1,â€¦,v_k} is linearly independent.

b. Find the dimension of Sp{v_1,â€¦,v_k,w}.

My answer:

a.

According to the information provided, the equation

x_1Â·v_1+â€¦x_kÂ·v_k=u has a single solution.

That means that for the solution to hold, all variables in this equation must have a fixed value, with no free variables. If u is the zero vector, each one of the variables must take on the value 0 for the single solution to hold: it is a given that there is a single solution only, so in this case it must be the trivial solution, as the trivial solution always exists in a homogeneous system. If the trivial solution is the only solution, then v_1 to v_k is linearly independent.

b.

per definition {v_1,â€¦,v_k} spans Sp{v_1,â€¦,v_k}, and it was shown that {v_1,â€¦,v_k} is linearly independent, so {v_1,â€¦,v_k} is a basis with dimension k. Sp{v_1,â€¦,v_k,w} has an additional member, w, which according to the information given is not a linear combination of {v_1,â€¦,v_k} (given that the equation x_1Â·v_1+â€¦+x_kÂ·v_k=w has no solution). Therefore {v_1,â€¦,v_k,w} is a basis of Sp{v_1,â€¦,v_k,w}, which dimension k+1.

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