# Entrance exams

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For each c  [c lying on the real line],define a function T_c:(R^4)–>(R^4) by

T_c((x1,x2,x3,x4)):=(([1+c]x1,x2+cx3,x3+cx2,[1+c]x4)).

For every c lying on the real line,find the dimension of the null space of T_c.

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Given: A linear transformation as follows:-

T_c: R4->R4  such that T_c(x1,x2,x3,x4)=([1+c]x1,x2+cx3,x3+cx2,[1+c]x4)

c belongs to R

To find the Nullity of T or the Dimension of The null space of T.

The Transformation Matrix with respect to the cannonical bases {e1,e2,e3,e4} is,

1+c  0  0  0

0    1  c  0   = T

0    c  1  0

0    0  0  1+c

Basically to find dim[N(T)].

We can apply row operations to reduce T to its row-reduced echelon form(RREF) in order to find its rank. The RREF of T is:

1+c  0    0      0

0    1    c       0   = RREF(T)

0    0  1-c^2  0

0    0    0    1+c

Now we know that the no.of non zero rows in the RREF of a matrix gives us the rank of the matrix i.e., Rank(T)=Rank of the Linear Transformation is clearly dependent on the values of c belonging to R.

Hence Rank(T)= 1, if c=-1

= 3, if c=1

=4, otherwise.

Now according to RANK NULLITY THEOREM we know that, nullity=n(no.of columns)- rank i.e.,

nullity(T)=dim(N[T])=3,if c=-1

=1,if c=1

=0,otherwise.

Note that when the nullity of the linear transformation is 0 this would imply that no other vector apart form the null vector 0 spans the Null Space of T.
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