# Which one of these two statements is wrong ?

#### Yankel

##### Active member
a. every set of n linearly independent vectors are basis of a vector space with dimension n

b. An invertible matrix is necessarily diagonalizable

they both seems wrong to me, but only one suppose to be.

'a' sounds wrong because the n vectors must also span the vector space

and 'b' because I don't see the relation between invertible and diagonalizable

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
'a' sounds wrong because the n vectors must also span the vector space
Which vector space?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
a. every set of n linearly independent vectors are basis of a vector space with dimension n

b. An invertible matrix is necessarily diagonalizable

they both seems wrong to me, but only one suppose to be.

'a' sounds wrong because the n vectors must also span the vector space

and 'b' because I don't see the relation between invertible and diagonalizable
n linearly independent vectors span a vector space with n dimensions.
Since the vector space has only n dimensions, those vectors necessarily must span that vector space.

For 'b' you should try to find a counter example.
Can you think of a matrix that is not diagonalizable, but still has a non-zero determinant?

#### jakncoke

##### Active member
to extend serena statement.

If D is a diagnizable matrix then there exists an invertible nxn matrix P
such that D = $P^{-1}AP$ where A is the matrix with the eigen values of D on its diagonals.

Now take D = $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$. whose eigen values are 1,1. Now if D is diagnoziable there would exist a nxn invertible matrix P such that

D = $P^{-1}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}P$, now 4 $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is just the identity matrix.
so D = $P^{-1}P = I$, since $D \not = I$, it is clear that D is not diagnizable.

Yet D is invertible Det(D) = 1.

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