Help on these linear algebra based questions. Let a, b be real numbers. Consider

Question

Help on these linear algebra based questions.

Let a, b be real numbers. Consider the equation z = ax + by. Prove that there are two 3-vectors v_1, v_2 such that the set of points [x,y,z] satisfying the equation is exactly the set of linear combinations of v_1 and v_2. Let a, b, c be real numbers. Consider the equation z = ax + by + c. Prove that there are three 3-vectors v_0, v_1, v_2 such that the set of points [x,y,z] satisfying the equation is exactly {v_0 + alpha_1 v_1 + alpha_2 v_2 : alpha_1 is element of real numbers set, alpha_2 is element of real numbers set}

Explanation / Answer

3.8.4)

The vectors v1 = e1 e2, v2 = e2 e3 are in V.

Let’s show that {v1, v2} is a basis of V.

Step (i): To check spanning, let v = (ax,by) be an arbitary vector in V.

We want to find scalars c1, c2 such that v = c1v1 + c2v2

Since z=ax+by,

we have v = xv1 + (x + y)v2 ,

so c1 = ax, c2 = b(x + y) do the job.

This shows that S spans V.

Step (ii): To check linear independence, we suppose there are scalars c1, c2 such that c1v1 + c2v2 = 0.

Writing both sides out in terms of components, this means (c1, c2 c1) = (0, 0),

which amounts to the equations c1 = 0 c2 c1 = 0 .

The only solution to these equation is c1 = c2 = 0. This shows that {v1, v2,} is linearly independent.

We have now shown that {v1, v2} is a basis of V.

This means that every vector v V can be uniquely written as v = c1v1 + c2v2 = (c1, c2 c1).

=ax+b(x+y)

=(ax,b(x+y)-ax).

3.8.5)

The vectors v1 = e1 e2, v2 = e2 e3, v3 = e3 e4 are in V.

Let’s show that {v1, v2, v3} is a basis of V.

Step (i): To check spanning, let v = (ax, by, c) be an arbitary vector in V.

We want to find scalars c1, c2, c3 such that v = c1v1 + c2v2 + c3v3.

Since z =ax+by+c, we have v = xv1 + (x + y)v2 + (x + y + z)v3,

so c1 = ax, c2 =b( x + y), c3 =c( x + y + z) do the job.

This shows that S spans V.

Step (ii): To check linear independence, we suppose there are scalars c1, c2, c3

such that c1v1 + c2v2 + c3v3 = 0.

Writing both sides out in terms of components, this means (c1, c2 c1, c3 c2, c3) = (0, 0, 0, 0), which amounts to the equations c1 = 0 c2 c1 = 0 c3 c2 = 0 c3 = 0.

The only solution to these equation is c1 = c2 = c3 = 0.

This shows that {v1, v2, v3} is linearly independent.

We have now shown that {v1, v2, v3} is a basis of V.

This means that every vector v V can be uniquely written as v = c1v1 + c2v2 + c3v3 = (c1, c2 c1, c3 c2, c3).

V=ax+b(x+y)+c(x+y+z)

   =(ax,b(x+y)-ax,c(x+y+z)-b(x+y),-c(x+y+z).

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