Proofs with Linearly independent vectors, Invertible matrices,

Your Problem

Suppose that  u₁, u₂, . . . , u_t are vectors in $C^n$ (complex number) which are linearly independent.

    (a) Also suppose that “M” is an n x n   matrix that is invertible.  Show that  Mu₁, Mu₂, . . . , Mu_t  are linearly independent vectors.

(b) This isn’t true when “M” is not invertible.  Find a counterexample, as follows.  Give an example of a 2 x 2 matrix “M” and two independent vectors  u₁, u₂  Є  $C^2$ so that Mu₁ and Mu₂  are dependent.

My Solution (a)

The concept in part (a) is quite easy, it is just a little technical to write it down correctly.

(a)

The vectors u₁, u₂, . . . , u_t   are linearly independent vectors if and only if the vector equation

a₁u₁ + a₂u₂ + . . . + a_t u_t = 0

has only the trivial solution, that is  

a₁ = a₂ = … = a_t = 0 .

Now consider the equation,   

a₁ (Mu₁) + a₂ (Mu₂)  + . . . + a_t (Mu_t) =0                (1)

We need to show the only solution for this is 

a₁ = a₂ = … = a_t = 0.

This will show that the vectors Mu₁, Mu₂, . . . , Mu_t  are linearly independent vectors.

Firstly, it is true that      a₁ (Mu₁) ₌ M (a₁ u₁)       

  (see appendix result 1)

Then the equation (1)  becomes

M (a₁ u₁) + M (a₂ u₂) + … + M (a_t u_t) = 0                          (2)

Then  factorizing we have

M ( (a₁ u₁) +  (a₂ u₂) + … + (a_t u_t) ) = 0                     (3)

(see appendix for proof of this)

Now since M is invertible it means that the inverse exists, so we can multiply both side by M^-1. The right side is just zero again, and M^-1 x M = I (the identity matrix)

M^-1 x M ( (a₁ u₁) + (a₂ u₂) + … + (a_t u_t) ) = M^-1 x 0

Therefore,

 ( (a₁ u₁) +  (a₂ u₂) + … + (a_t u_t) ) = 0

Dropping the brackets as we do not need them anymore, they are just grouping symbols,

a₁ u₁ + a₂ u₂ + … + a_t u_t  = 0

But since the vectors u₁, u₂, . . . , u_t   are linearly independent vectors by assumption then the only solution to this equation is

a₁ = a₂ = … = a_t = 0

So we are done since we have shown that the only solution to the equation (1) is the trivial one, and hence the vectors

Mu₁, Mu₂, . . . , Mu_t  are linearly independent vectors as required.

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My Solution (b)  

Appendix:

Result 1:

If a is a number and M = (m_ij) is an nxn matrix and u = (u_j) is an n-vector then we show my the basic definitions of multiplication between scalars, vectors and numbers that

a (Mu)  =  M (a u)                                               (*)

The i^th entry of the Left Hand Side vector

=  a (∑ⁿ_j=1  m­­­­_ij u_j)

=  ∑ⁿ_j=1  a m­­­­_ij u_j

=  ∑ⁿ_j=1  m­­­­_ij  (au_j ) = right hand side

but this is the i^th  entry of the right hand side, and hence we have proven (*)

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Result 2                         

If M = (m_ij) is an nxn matrix and u = (u_j)  v = (v_j)  are n-vectors then we show my the basic definitions  vectors and matrix operations that

M(u + v)  =  Mu + Mv                                (**)

           The i^th entry of the Left Hand Side vector in (**)

= ∑ⁿ_j=1  m­­­­_ij  (u_j + v_j )

= ∑ⁿ_j=1  (m­­­­_ij u_j + m­­­­_ij v_j )       (by basic rules of complex numbers multiplication)

          = ∑ⁿ_j=1  m­­­­_ij u_j + ∑ⁿ_j=1  m­­­­_ij v_j

          = the i^th entry of Mu + the i^th entry of Mv

          = the i^th entry right hand side in (**)

and so we are done and (**) is proven.

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[Quadratic Function] Sketch the graph of the quadratic function $f(x)=-3x^2-6x-5$. Give the intercepts, the vertex, the axis, the domain, and the range.

QUESTION:
Sketch the graph of the quadratic function $f(x)=-3x^2-6x-5$. Give the intercepts, the vertex, the axis, the domain, and the range.

SOLUTION.


Therefore,